Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this document? Why not share!

- Project application of derivatives ... by TaraRocheleDaugherty 11654 views
- Application of derivative by Amiraj College Of... 1351 views
- Application Of Derivative In Real L... by Afnanul Hassan 1997 views
- The Application of Derivatives by divaprincess09 12541 views
- Applications of Derivatives by Iram Khan 35963 views

7,795 views

Published on

No Downloads

Total views

7,795

On SlideShare

0

From Embeds

0

Number of Embeds

20

Shares

0

Downloads

173

Comments

0

Likes

2

No embeds

No notes for slide

- 1. Differentiation and application of derivativesIncreasing and Decreasing FunctionsThis section explains how derivative can be used to check whether afunction is increasing, decreasing or neither increasing nor decreasingin its domain.Let f be a function defined on an interval I and let x1 and x2 be any twopoints on I.(i) f is said to be increasing in the interval I,Example:DefineThe graph of the function is as follows:f(x) is increasing, because
- 2. Since 2 - 1 < 3 - 1(ii) f is said to strictly increasing in the interval I ifFor example,Let x1 < x2f(x) is strictly increasing function.(iii) f(x) is said to be decreasing function if for x1, x2 IFor example,
- 3. f(x) = 1 - x for 0 < x < 1= 0 for 0 x <2= 2 - x for x 2is a decreasing function.(iv) f (x) is said to be strictly decreasing on an interval I if forFor example, is strictly decreasing function.Theorem 1:Let f be continuous on [a, b] and differentiable on the open interval (a,b). Then(a) f is increasing on [a, b] if f (x) > 0 for each x (a, b)(b) f is decreasing on [a, b] if f (x) < 0 for each x (a, b)This theorem can be proved by using Mean Value Theorem. We shall
- 4. prove the theorem after learning Mean Value Theorem.This theorem is applied in various problems to check whether afunction is increasing or decreasing.Working Rule to Check Whether a Differentable Function is Increasingor Decresing(1) Let the given function be f (x) on the real number line R.(2) Differentiate the function f(x) with respect to x and equate it to zeroi.e., put f (x) = 0. Solve for x. These values of x which satisfy f (x) = 0are called Critical values of the function(3) Arrange these Critical values in ascending order and partition thedomain of f (x) into various intervals, using the Critical values.(4) Check the sign of f (x) in each open intervals.(5) If f (x) > 0 in a particular interval, then the function is increasing inthat particular interval.If f (x) < 0 in a particular interval, then the function is decreasing inthat particular interval.Example:Find the intervals on which the function(a) increasing (b) decreasingDifferentiating the function, we have
- 5. The critical values in ascending order are -1, 1. We divide the Realnumbers into the intervals= - veSince f (x) < 0, the function is decreasing in the interval .
- 6. Since f (x) > 0 in the interval (-1, 1), the function is increasing in thisinterval.= - veSince f (x) < 0, f(x) is decreasing in the interval (1, )Maxima and MinimaA function f(x) is said to have a local maximum at x = a, if is aneighbourhood I of a, such thatf(a) f(x) for all x IThe number f(a) is called the local maximum of f(x). The point a is calledthe point of maximum.
- 7. Note that when a is the point of local maxima, f(x) is increasing for allvalues of x < a and f (x) is decreasing for all values of x > a in the giveninterval.At x = a, the function ceases to increase.A function f(x) is said to have a local minimum at x = a, if is aneighbourhood I of a, such thatf(a) f(x) for all x IHere, f(a) is called the local minimum of f(x). The point a is called thepoint of minimum.Note that, when a is a point of local minimum f (x) is decreasing for all x< a and f (x) is increasing for all x > a in the given interval. At x = a, thefunction ceases to decrease.If f(a) is either a maximum value or a minimum value of f in an interval I,then f is said to have an extreme value in I and the point a is called theextreme point.Monotonic Function
- 8. A function is said to be monotonic if it is either increasing or decreasingbut not both in a given interval.Consider the functionThe given function is increasing function on R. Therefore it is amonotonic function in [0,1]. It has its minimum value at x = 0 which isequal to f (0) =1, has a maximum value at x = 1, which is equal to f (1) =4.Here we state a more general result that, Every monotonic functionassumes its maximum or minimum values at the end points of itsdomain of definition.Note that every continuous function on a closed interval has amaximum and a minimum value.Theorem 2:(First Derivative Test)Let f (x) be a real valued differentiable function. Let a be a point on aninterval I such that f (a) = 0.(a) a is a local maxima of the function f (x) ifi) f (a) = 0ii) f (x) changes sign from positive to negative as x increases through a.That is, f (x) > 0 for x < a andf (x) < 0 for x > a(b) a is a point of local minima of the function f (x) ifi) f (a) = 0
- 9. ii) f (x) changes sign from negative to positive as x increases through a.That is, f (x) < 0 for x < af (x) > 0 for x > aWorking Rule for Finding Extremum Values Using First Derivative TestLet f (x) be the real valued differentiable function.Step 1: Find f (x)Step 2: Solve f (x) = 0 to get the critical values for f (x). Let these valuesbe a, b, c. These are the points of maxima or minima.Arrange these values in ascending order.Step 3: Check the sign of f(x) in the immediate neighbourhood of eachcritical value.Step 4: Let us take the critical value x= a. Find the sign of f (x) forvalues of x slightly less than a and for values slightlygreater than a.(i) If the sign of f (x) changes from positive to negative as x increasesthrough a, then f (a) is a local maximum value.(ii) If the sign of f (x) changes from negative to positive as x increasesthrough a, then f (a) is local minimum value.(iii) If the sign of f (x) does not change as x increases through a, then f(a) is neither a local maximum value not a minimum value. In this case x= a is called a point of inflection.Example:Find the local maxima or local minima, if any, for the following functionusing first derivative test
- 10. f (x) = x3 - 6x2 + 9x + 15Solution:f (x) = x3 - 6x2 + 9x + 15f (x) = 3x2 -12x + 9= 3(x2- 4x + 3)= 3 (x - 1) (x - 3)Thus x = 1 and x = 3 are the only points which could be the points oflocal maxima or local minima.Let us examine for x=1When x<1 (slightly less than 1)f (x) = 3 (x - 1) (x - 3)= (+ ve) (- ve) (- ve)= + veWhen x >1 (slightly greater than 1)f (x) = 3 (x -1) (x - 3)= (+ ve) (+ ve) (- ve)= - ve
- 11. The sign of f (x) changes from +ve to -ve as x increases through 1. x = 1 is a point of local maxima andf (1) = 13 - 6 (1)2 + 9 (1) +15= 1- 6 + 9 + 15 =19 is local maximum value.Similarly, it can be examined that f (x) changes its sign from negative topositive as x increases through the point x = 3. x = 3 is a point of minima and the minimum value isf (3) = (3)3- 6 (3)2+ 9(3) + 15= 15Theorem 3: (Second Derivative Test)Let f be a differentiable function on an interval I and let a I. Let f "(a) becontinuous at a. Theni) a is a point of local maxima if f (a) = 0 and f "(a) < 0ii) a is a point of local minima if f (a) = 0 and f "(a) > 0iii) The test fails if f (a) = 0 and f "(a) = 0. In this case we have to go backto the first derivative test to find whether a is a point of maxima,minima or a point of inflexion.Working Rule to Determine the Local Extremum Using SecondDerivative TestStep 1For a differentiable function f (x), find f (x). Equate it to zero. Solve theequation f (x) = 0 to get the Critical values of f (x).
- 12. Step 2For a particular Critical value x = a, find f "(a)(i) If f (a) < 0 then f (x) has a local maxima at x = a and f (a) is themaximum value.(ii) If f (a) > 0 then f (x) has a local minima at x = a and f (a) is theminimum value.(iii) If f (a) = 0 or , the test fails and the first derivative test has to beapplied to study the nature of f(a).Example:Find the local maxima and local minima of the function f (x) = 2x3 - 21x2+36x - 20. Find also the local maximum and local minimum values.Solution:f (x) = 6x2 - 42x + 36f (x) = 0 x = 1 and x = 6 are the critical valuesf (x) =12x - 42If x =1, f (1) =12 - 42 = - 30 < 0 x =1 is a point of local maxima of f (x).
- 13. Maximum value = 2(1)3 - 21(1)2 + 36(1) - 20 = -3If x = 6, f (6) = 72 - 42 = 30 > 0 x = 6 is a point of local minima of f (x)Minimum value = 2(6)3 - 21 (6)2 + 36 (6)- 20= -128Absolute Maximum and Absolute Minimum Value of a FunctionLet f (x) be a real valued function with its domain D.(i) f(x) is said to have absolute maximum value at x = a if f(a) f(x) for allx D.(ii) f(x) is said to have absolute minimum value at x = a if f(a) f(x) for allx D.The following points are to be noted carefully with the help of thediagram.Let y = f (x) be the function defined on (a, b) in the graph.(i) f (x) has local maximum values at
- 14. x = a1, a3, a5, a7(ii) f (x) has local minimum values atx = a2, a4, a6, a8(iii) Note that, between two local maximum values, there is a localminimum value and vice versa.(iv) The absolute maximum value of the function is f(a7)and absoluteminimum value is f(a).(v) A local minimum value may be greater than a local maximum value.Clearly local minimum at a6 is greater than the local maximum at a1.Theorem 4Let f be a continuous function on an interval I = [a, b]. Then, f has theabsolute maximum value and f attains it at least once in I. Also, f has theabsolute minimum value and attains it at least once in I.Theorem 5Let f be a differentiable function on I and let x0 be any interior point of I.Then(a) If f attains its absolute maximum value at x0, then f (x0)= 0(b) If f attains its absolute minimum value at x0, then f (x0) = 0.In view of the above theorems, we state the following rule for finding theabsolute maximum or absolute minimum values of a function in a giveninterval.Step 1: Find all the points where f takes the value zero.Seep 2: Take the end points of the interval.
- 15. Step 3: At all the points calculate the values of f. Step 4: Take the maximum and minimum values of f out of the values calculated in step 3. These will be the absolute maximum or absolute minimum values. Rolles Theorem and Mean Value Theorem Rolles TheoremLet f be a real valued function in [a,b] such that f is continuous in [a,b]. f is differentiable in (a,b).Geometrical meaningLet A (a,f (a)) and B (b,f (b)) be two points on the graph of f (x) such that f(a)= f(b), then c (a, b) such that the tangent at P(c, f(c)) is parallel to x -axis.
- 16. Note 1: We cannot obtain c if any one of the conditions of Rolles theoremare not satisfied.Note 2: The value of c need not be unique.Example:Verify Rolles theorem for the functionf (x) = x2 - 8x + 12 on (2, 6)Since a polynomial function is continuous and differentiable everywhere f(x) is differentiable and continuous (i) and (ii) conditions of Rolles theoremis satisfied.f (2) = 22 - 8 (2) + 12 = 0f (6) = 36 - 48 + 12 = 0Therefore (iii) condition is satisfied. Rolles theorem is applicable for the given function f (x). There must exist c (2, 6) such that f (c) = 0f (x) = 2x - 8 Rolles theorem is verified.Working Rule for Verifying Rolles TheoremLet f (x) be a function defined on [a, b].
- 17. Step 1:Show that the function is continuous in the given interval. Some knownstandard functions which are continuous, can be mentioned directly.Step 2:Differentiate f (x) and examine if f (x) is defined at every point in the openinterval (a, b).Step 3:Check if f (a) = f (b)If all the above condition are satisfied, then Rolles theorem is applicableelse the Rolles theorem is not applicable.Langranges Mean Value TheoremTheorem 7:Let f be real valued function in [a,b] such that, f is continuous in [a,b]. f is differentiable in (a,b).Geometrical meaningLet A (a,f (a)) and B (b,f (b)) be two points on f (x), then
- 18. Note: The value of c obtained need not be unique.Working Rule to Verify Lagranges Mean Value Theorem(Mean Value Theorem)Step 1:Show the function f (x) is continuous on the closed interval [a, b].Step 2:Find f (x) and examine if it is defined at every point on the open interval (a,b). If f (x) is defined for all x (a, b), then the function is differentiable.Step 3:If the above condition are satisfied, then Mean Value Theorem is applicable.Step 4:If Mean value theorem is applicable, solve the equation
- 19. Show that one of the roots lie in the open interval (a, b). This verifies theMean Value Theorem.Example:Verify mean value theorem for the functionf (x) = (x - 4) (x - 6) (x - 8) in [4,10]Step1:We know that every polynomial function is continuous and product ofcontinues functions are continuous. f (x), being product of polynomials ofdegree 1, is a continuous function in [4,10].Step 2:f (x) = (x - 6) (x - 4) + (x - 4) (x - 8) + (x - 6) (x - 8)f (x)= (x2 -10x + 24) + (x2 - 12x + 32)+ (x2 - 14x + 48)= 3x2 - 36x + 104f (x) is defined for all values on the interval (4,10). f (x) is differentiable.Step 3:Since both the condition are satisfied, Mean Value Theorem is applicable.There exist c (4, 10) such that
- 20. f (4) = 0f (c) = 3c2 - 36c + 104Substituting these values in (1), we haveSince 8 (4,10), Mean Value Theorem is satisfied. Approximations by Differentials Let y = f (x) be a differentiable function of x, errors in x and y are denoted by x and y, we have Error in y = f (x) x.
- 21. Note 1:Note 2:Note 3:y and dy are not usually the same and dy is the approximate valueof y.Note 4: dx and dy are called the differentials of x and y respectively.Example:Find the approximate value of fourth root of 82 upto 3 decimalplaces.
- 22. Solution:Let y = f (x) = x1/4Let x = 81, x =1. Taking these values we have Rate of Change of Quantity If a quantity y varies with respect to another quantity x satisfying some rule y = f(x), in other words if y is a function x, then represents the rate of change of y with respect to x For x = x0, dy/dx at x0 is called the rate of change of y with respect to x at x0. If y is a function of t x is a function of t That is if x = f (t) and y = g(t)
- 23. We know thatThe rate of change of y with respect to x in this case, can befound by finding out the rate of change of y with respect to tand that of x with respect to t.Example:A balloon which always remain spherical is being inflated bypumping in 900 cubic centimetres of gas per second. Finalthe rate at which the radius of the balloon increasing whenthe radius is 15cm.Let r and V be the variables for radius and volume of theballoon respectively.We have to findFor a sphereDifferentiating both sides with respect to t, we have
- 24. Tangents and NormalsNote 1: If m = 0 the tangent is parallel to x-axis.Note 3: Equation of the tangent at (x1,y1) isorNote 4: Equation of the normal at (x1,y1) isorExample:
- 25. Find the equation of the tangent to the curve which isparallel to the line 4x - 2y + 5 = 0.Solution:Since the tangent to the given curve is parallel to the line4x - 2y + 5 = 0Slope of the tangent = Slope of the given line
- 26. On simplification, the equation is 48 x - 24y = 23.Curve SketchingThe graph of a given function gives a visual presentation of thebehaviour of the function involving a study of symmetry, rise andfall, region of existence, passage through specific points etc.The following points are helpful to trace the curve.Symmetrya) Symmetry about x - axisIf the equation of the curve remains unaltered when y is replacedby -y, then the curve is symmetrical about x-axis.b) Symmetry about y-axisIf the equation of the curve remains unaltered when x is replacedby -x then the curve is symmetrical about y-axis.c) Symmetry about y = xIf the equation of the curve remains unaltered if x and y areinterchanged, then the curve is symmetrical about y = x.d) Symmetry about y = -xIf x and y are replaced by -y and -x and the equation of the curve isunaltered, then the curve is symmetrical about the line y = -x.e) Symmetry in opposite quadrantsIf the equation of the curve is unaltered, when x and y are replaced
- 27. by -x and -y, then it is symmetrical in opposite quadrants.Example:Sketch the curvey = - sin 2x ….(1)Solution:Symmetry(a) By replacing y by -y, the equation (1) is altered, therefore thecurve is not symmetrical about x-axis.(b) By replacing x by -x, the equation of the curve is altered,therefore the curve is not symmetrical about y-axis.(c) Replace x and y by -x and -y respectively in the equation y = -sin 2x, the equation of the curve remain unaltered. Therefore thecurve is symmetrical in opposite quadrants.Passage through originPut x = 0, y = -sin2x = 0. This implies (0, 0) is a point on the curve orthe curve passes through the origin.Points of intersectionThe points of intersection with x-axis is determined by letting y = 0.Putting y = 0, - sin 2x = 0 This implies the curve intersects the x-axis at the points where
- 28. The points at which the tangent is parallel to x-axes are determinedby solving.y = - sin 2x The tangent is parallel to x- axis atWe know that sin x is a periodic function with 2, that is
- 29. sin(2 + x) = sinx The pattern of the curves repeats at an interval .Therefore it is sufficient to sketch the curve form 0 to .Determination of Few Pointwith this information, we sketch the graph as show .Summary
- 30. Let y = f(x) be a smooth curve and P(x,y) be a point on the curve. Equation of the tangent at (x1, y1) in the curve y = f (x) isy - y1 Equation of the normal at (x1, y1) in the curve y = f (x) is
- 31. If m = 0 the tangent at (x1, y1) is parallel to x-axis. Angle of intersection of the curves is the acute angle betweentheir tangents at the point of intersection.Let y = f1(x) and y = f2(x) be two curves intersecting at P. If m1 = m2 the curves touch each other. If m1m2 = -1 the curves are orthogonal. The condition for the function f(x) to be increasing at x = a iff(a) >0. The condition for f(x) to be decreasing at x = a if f(a) < 0.A function f(x) is said to be strictly increasing at x = a if f(x) > f(a)whenever x > a in the neighbour hood of a.
- 32. A function f(x) is said to be strictly decreasing at x = a if f(x) < f(a)whenever x>a in the neighbour hood of a If f(a) = 0 then x = a is called a stationary point.If f(a)=0, x=a is called the stationary point and the tangent at(a,f(a)) is parallel to x - axis. A function f(x) is said to have a maximum at x = a if
- 33. The conditions for y = f(x) to have a maximum at x = a are A function f(x) is said to have a minimum at x = a The conditions for y = f(x) to have a minimum at x = a arex = a is called a point of inflexion.
- 34. Rolles theorem: If a function f(x) is such that(i) f (x) is continuous on [a,b](ii) f (x) is differentiable on (a,b) and(iii) f (a) = f (b)Geometrical interpretation of Rolles theoremLet AB be the graph of y = f(x) such that A = (a,f(a)) and B = (b,f(b))(i) f (x) is continuous between A and B.(ii) f (x) has derivative between A and B i.e., there is a unqiuetangent at every point between A and B.(iii) f (a) = f(b)at C is parallel to x-axis.
- 35. Langranges Mean Value theorem: If a function f (x) is such that(i) f (x) is continuous on [a,b] and(ii) f (x) is differentiable on (a,b) thenGeometrical interpretation of Lagranges Mean value theorem.Let A(a, f(a)) and B(b, f(b)) be two points on the curve y = f (x) suchthat(i) f (x) is continuous between A and B.(ii) f (x) is differentiable between A and Bi.e., there is a unique tangent at every point between A and B.then there exists a point C(c, f(c)) between A and B such that thetangent at C is parallel to the chord AB.
- 36. Note that the point C is not unique.Relation between y and dyLet A(x, y) and B(x + x, y + y) be two neighbouring points on thecurve y = f(x).Let dx and dy be the differentiables of x and y respectively.
- 37. AC = x = dxBC = yDC = dydy = f(x)xy - dy = BC - CD = BD The differential dy and the increment y may not be equal. Generally the following points are examined for tracing thecurves
- 38. (i) Symmetry about x - axis(ii) Symmetry about y - axis(iii) Symmetry in opposite quadrants(iv) Symmetry about the line y = x(v) Passage through specific points(vi) Points of intersection with the axes(vii) Regions where the curve is increasing or decreasing(viii) Region of existence.PROBLEMSTo find the equation of the tangent line ofFirst, we find the derivative:
- 39. The equation of the tangent line is:To find the equation of the tangent line ofFirst, we find the derivative:
- 40. The equation of the tangent line is:To find the equation of the tangent line ofFirst, we find the derivative:
- 41. The equation of the tangent line is:To find the equation of the tangent line ofFirst, we find the derivative:The equation of the tangent line is: y - 0.333 = -0.667 (x - 0.5)To find the equation of the tangent line ofFirst, we find the derivative:
- 42. The equation of the tangent line is: y - 1 = -1 (x - 1)Problem:Find the inverse ofSolution:Frist, we differentiateThe derivative of f -1 is:
- 43. The tangent line at (-1, 0) is:Problem:Find the inverse ofSolution:Frist, we differentiateThe derivative of f -1 is:
- 44. The tangent line at (7, 1) is:Problem:Using logarithmic differentiation, differentiate:Solution:Problem:Using logarithmic differentiation, differentiate:Solution:Problem:Using logarithmic differentiation, differentiate:
- 45. Solution:Problem: Using differentials, approximate the expressionSolution:We letHence, x = 2 and y = 5.Differentiating, we obtainSubstituting, we get
- 46. **Problem: Using differentials, approximate the expressionSolution:We letHence, x = -.0015926... and y = 0.Differentiating, we obtainSubstituting, we get**Problem: Using differentials, approximate the expressionSolution:We let
- 47. Hence, x = .035398... and y = 1.Differentiating, we obtainSubstituting, we get**Problem: Using differentials, approximate the expressionSolution:We letHence, x = 0.05 and y = 0.Differentiating, we obtain
- 48. Substituting, we get**Problem: Using differentials, approximate the expressionSolution:We letHence, x = -0.1 and y = 9.Differentiating, we obtainSubstituting, we get
- 49. **Problem: Using differentials, approximate the expressionSolution:We letHence, x = 0.05 and y = /4.Differentiating, we obtainSubstituting, we getTo find the absolute maxima and minima of the function
- 50. we first take the derivativeand find the roots of the derivative which are: x = 0.63486... and x = 2.50672....These are the only critical points of f. We consider the following table ofthe endpoints and the critical points of f: x f(x) 0 1 3.14159... -1 0.63486... 1.76017... 2.50672... -1.76017...from which we see that the absolute maximum occurs at x = 0.63486... andis 1.76017... and the absolute minimum occurs at x = 2.50672... and is -1.76017... .To find the absolute maxima and minima of the functionwe first take the derivativeand find the roots of the derivative which are: x = -2 and x = 0.These are the only critical points of f. We consider the following table ofthe endpoints and the critical points of f: x f(x)
- 51. -5 0.16844... 1 2.71828... -2 0.54134... 0 0from which we see that the absolute maximum occurs at x = 1 and is2.71828... and the absolute minimum occurs at x = 0 and is 0.To find the absolute maxima and minima of the functionwe first take the derivativeand find the root of the derivative which is: x = -1.This is the only critical point of f. We consider the following table of theendpoints and the critical points of f: x f(x) -4 2.48490... 3 2.94443... -1 1.09861...from which we see that the absolute maximum occurs at x = 3 and is2.94443... and the absolute minimum occurs at x = -1 and is 1.09861... .Problem: Using LHospitals Rule, evaluate the limitSolution:Since
- 52. we can apply LHospitals Rule:Problem: Using LHospitals Rule, evaluate the limitSolution:Sincewe can apply LHospitals Rule:Problem: Using LHospitals Rule, evaluate the limitSolution:Sincewe can apply LHospitals Rule:
- 53. To apply LHospitals Rule again, we need to check:Problem: Using LHospitals Rule, evaluate the limitSolution:Sincewe can apply LHospitals Rule:
- 54. Problem: Using LHospitals Rule, evaluate the limitSolution:Sincewe can apply LHospitals Rule:
- 55. Problem: Using LHospitals Rule, evaluate the limitSolution:Sincewe can apply LHospitals Rule:
- 56. To apply LHospitals Rule again, we need to check:To apply the Mean Value Theorem to the function
- 57. we first calculate the quotientNext, we take the derivativeand equate it to the result of the calculation above:We solve this equation to get two answers: x = 0.7704383989 and x = 2.371154255.**To apply the Mean Value Theorem to the functionwe first calculate the quotientNext, we take the derivativeand equate it to the result of the calculation above:We solve this equation to get x = -0.9012084345.
- 58. **To apply the Mean Value Theorem to the functionwe first calculate the quotientNext, we take the derivativeand equate it to the result of the calculation above:We solve this equation to get three answers: x = 1.109744819, x = 2.403237586 and 4.022404459. We are given the function First, we find the derivative: We set the derivative equal to 0 and solve: Since the domain of f is the same as the domain of f, -2 and 0 are the only critical numbers of f. Testing:
- 59. x < -2 f(-10) = 80 e- 10 f is increasing -2 < x < 0 f(-1) = - e- 1 f is decreasing x>0 f(1) = 3 e f is increasing By the First Derivative Test, x = - 2 is a local minimum and x = 0 is a local maximum. We are given the function First, we find the derivative: We set the derivative equal to 0 and solve: Since f is not defined at - 3 and 3 which are in the domain of f, -3, 0 and 3 are the critical numbers of f. Testing: x < -3 f(-10) = -20 f is decreasing -3 < x < 0 f(-1) = 2 f is increasing 0<x<3 f(1) = -2 f is decreasing x>3 f(10) = 20 f is increasing
- 60. By the First Derivative Test, x = -3 and x = 3 are local minima and x = 0 is a local maximum. We are given the function First, we find the derivative: We set the derivative equal to 0 and solve: Since the domain of f is the same as the domain of f, 0 is the only critical number of f. Testing: x < 0 f(-1) = -1 f is decreasing x>0 f(1) = 1 f is increasing By the First Derivative Test, x = 0 is a local minimum. We are given the function First, we find the first and second derivatives:
- 61. Clearly, there is no solution for f(x) = 0. Note that - 3 and 3 are in the domain of f but are not in the the domain of f; hence, x = -3 and x = 3 are the only possible candidates for inflection points. Testing: x < -3 f(-5) = 2 f is concave upward -3 < x < 3 f(0) = -2 f is concave downward x>3 f(1) = 2 f is concave upward Therefore, x = -3 and x = 3 are inflection points. We are given the function First, we find the first and second derivatives: We set the second derivative equal to 0 and solve:
- 62. Since the domain of f is the same as the domain of f, x = -3.41.. and x = -0.58.. are the only possible candidates for inflection points. Testing: x < -3.41.. f(-4) = 0.036.. f is concave upward -3.41.. < x < -0.58.. f(-1) = -0.36.. f is concave downward x > -0.58.. f(0) = 2 f is concave upward Therefore, x = -3.41.. and x = -0.58.. are inflection points. We are given the function First, we find the first and second derivatives: Clearly, there is no solution for f(x) = 0. Note that the domain of f is [ 0, 4 ] and the domain of f is ( 0, 4 ); hence, there are no inflection points. Testing: 0 < x < 4 f(1) = -0.769.. f is concave downward( Ncert) Ex.6.1
- 63. qQuestion 2:The volume of a cube is increasing at the rate of 8 cm3/sec.HHow fast is the surface area increasing when the length of an edge is12 cm? Answer 2: V= X3 ( x is a side) , S = 6X2 ⇨ dv/dt (8 cm3/s)= 3x2.dx/dt Put dx/dt in dS/dt = 32/x cm2/sec, at x=12 cm will be 8/3 cm2/sec. Question 4: An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long? Answer 4: V= x3 , Rate of change in volume = dV/dt, , at x=10 is 900 cm3/sec. (∵ dx/dt=3 cm.) Question 5:A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing? Answer 5: A= r2 , dr/dt =5cm/sec. , dA/dt= 10 r=80 cm2/sec.(r=8) Question 7:The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle. Answer 7:(same as example)(a) dP/dt=2cm/min.(perimeter decreases) (b) dA/dt=2cm2/min.(shows area increases). Question 8:A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. Answer 8: V=4/3 r3 , dV/dt=900=4 r2dr/dt at r=15⇨ dr/dt=7/22 Question 10:A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
- 64. Answer 10: AB is ladder(hyp.)OA=4=x(base), OB=y(per.) so x2+y2=25Y=3(by pytha.) diff. W.r.t. t ⇨ dx/dt=.02m/sec. & dy/dt=-2/75(dec.)Question 13:A balloon, which always remains spherical, has a variablediameter Find the rate of change of its volume with respectto x. [Ans. dv/dx= 27 /8(2x+1)2 ]Question 14:Sand is pouring from a pipe at the rate of 12 cm3/s. Thefalling sand forms a cone on the ground in such a way that the heightof the cone is always one-sixth of the radius of the base. How fast isthe height of the sand cone increasing whenAnswer 14: put r=6h in Volume gives 12 h3 ⇨ dV/dt= 12 cm3/s. ⇨dh/dt= 1/(3 h2) =1/48 at h=4cm.Question 16:The total revenue in Rupees received from the sale of xunits of a product is given by Find the marginal revenue when x = 7.Answer 16: Marginal Revenue(MR) = dR/dx, at x=7 is Rs. 208.EX. 6.2 Question 5: Find the intervals in which the function f given byf(x)= 2x3 – 3x2 – 36x+7 is strictly increasing or decreasing.Answer 5: f is strictly ↑ if f’(x)>0 & strictly ↓ or f is strictly increasing on(-∞, -2)U(3, ∞) and decreasing on (-2, 3)Question 6:Find the intervals in which the following functions arestrictly increasing or decreasing: (a) x2 + 2x − 5 (b) 10 − 6x − 2x2(c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9x − x2 (e) (x + 1)3 (x − 3)3Answer 6: (a) f is strictly increasing on (-1, ∞) and decreasing on (-∞, -1) (b) f is strictly increasing on (-∞, -3/2) and decreasing on (-3/2, ∞)(c) f is strictly increasing on -2<x<-1 and decreasing on x<-2 & x>-1.(d) fis strictly increasing for x<-9/2 and decreasing for x>-9/2. (e) f’(x)=6(x+1)2(x-3)2(x-1) > & <0 for ↑ & ↓ , f is ↑ on (1, ∞) & ↓ on (-∞,1),(-1,1)
- 65. Question 7:Show that , is an increasingfunction of x throughout its domain.Answer 7: f’(x) = > 0 ⇨ f is increasing function of x ∀ x>-1Question 9: Prove that is an increasing functionof θ in .Answer 9: dy/dx= > 0 ⇨ cosѲ >0,[ )>0 as cos2ѲIs not >1] ∴ Ѳ ∊ (0, п/2)Question 11:Prove that the function f given by f(x) = x2 − x + 1 isneither strictly increasing nor strictly decreasing on (−1, 1).Answer 11: f’(x)=2(x-1/2), -1<x<1/2 ⇨ (x-1/2)<0⇨ f’(x)<0 and ½<x<1⇨ x-1/2>0 i.e., f’(x)>0, f’(X) does’nt have same sign in interval (-1,1).Question 12:Which of the following functions are strictly decreasingon ? (A) cos x (B) cos 2x (C) cos 3x (D) tan xAns. 12: A,B are ↓ on (o, п/2) , (C) is neither ↑ nor ↓ on (D) ↑ on (o, п/2)Question 13:On which of the following intervals is the function f givenby strictly decreasing?(A) (B) (C) (D) None of theseAnswer 13: f’(x) =100x99+cosx (A) 1 radian = 570 ∴ x∊ (0,1) means x liesin 1st quad. ∴ cosx >0⇨ 100x99+cosx>0 ⇨ f is ↑ . (B) п/2<x<п ⇨22/14<x<22/7 means 100x99 >100 & п/2<x<п ⇨ -1< cosx<0 or0>cosx>-1 ∴ 100x99+cosx > 100-1=99 ⇨ f is ↑. (C) f is ↑ (D) is correct.Question 14:Find the least value of a such that the function f given is strictly increasing on (1, 2).Answer 14: f’(x) = 2x+a , for f to be ↑ on [1,2], f’(x) should be +ve in[1,2]. At the end point x=1 2x+a=2+a which is +ve or zero i.e., 2+a≥0
- 66. At the end point x=2 2x+a=4+a which is +ve or zero i.e., 4+a≥0 if a≥-4∴ from above results we obtain the least value of a = -2.Question 15:Let I be any interval disjoint from (−1, 1). Prove that thefunction f given by is strictly increasing on I.Answer 15: f’(x) = ≥ 0 , is ↑ ∵ x2≥1≥0 , f is strictly increasing on I.Question 16:Prove that the function f given by f(x) = log sin x is strictlyincreasing on and strictly decreasing onAnswer 16: f’(x) = cotx , f is ↑ on & ↓ on as f’ is +ve & -ve.Question 17:Prove that the function f given by f(x) = log cos x is strictlydecreasing on and strictly increasing onEXTRA questionsQues. 1 Determine the value of x for which f(x) = xx , x>0 is increasingor decreasing. xlogxAnswer: Domain f =(0,∞) , f’(x) = e d/dx(xlogx)= xx (1+logex)For ↑ 1+logex>0 as xx >0 for x>0 ⇨ logex >-1 ⇨ x> e-1 , x∊ (1/e, ∞)Similarly f is ↓ on (0, 1/e)Ques. 2Find the intervals in which f(x) = sinx – cosx, 0< x< 2п, is strictlyincreasing or strictly decreasing .Answer: f’(x) = cosx+sinx= √2(1/√2 sinx + 1/√2 cosx)= √2 sin(x+п/4)0<x<2п ⇨ п/4<x+п/4<9п/4 ,we know that sin(x+п/4)>0 forп/4<x+п/4<п or 2п<x+п/4<9п/4 ⇨ sin(x+п/4)>0 for 0<x <3п/4 or7п/4<x<2п ⇨ f is strictly ↑ in (0,3п/4)U(7п/4,2п) & ↓ in [3п/4,7п/4]Question 3: Find local max. & mini. Value of f(x) =sin4x + cos4x,
- 67. x∊(0,п/2) . *Answer: f’(x)=0⇨-sin4x =0 ⇨ x=п/4, put f’’(x)>0, local mini.Value is ½]Question 4: Shoe that the rectangle of max. Area that can be inscribedin a circle of radius r is a square of side √2r.Answer: r be radius , Ѳ be the angle b/w side of rectangle & diameterA= 2rcosѲ.2rsinѲ, dA/dѲ=0 ⇨ cos2Ѳ=0 ⇨ Ѳ=п/4, show 2nd der. <0Sides are √2r,so it becomes square has max. Area.EX.6.3 Question 9:Find the point on the curve y = x3 − 11x + 5 at whichthe tangent is y = x − 11.Answer 9: dy/dx = 3x2-11 , slope of tangent (m)=1, after solving x=∓2⇨y=-13,-9 ,so(2,-9) will be the point ∵ (-2,-13) does not lie on curve.Question 13: Find points on the curve at which the tangentsare (i) parallel to x-axis (ii) parallel to y-axisAnswer 13: dy/dx = -16x/9y (i) dy/dx=0⇨x=0 ∴y=∓4 (ii) dy/dx=∞⇨y=0Question 15: Find the equation of the tangent line to thecurve y = x2 − 2x + 7 which is (a) parallel to the line 2x − y + 9 = 0(b) perpendicular to the line 5y − 15x = 13.Answer 15: dy/dx=2(x-1) (a) dy/dx = m= 2⇨ x=2 & y=7, eqn. Is 2x-y+3=0(b) dy/dx=m=3, but it is per. ∴ 2(x-1) = -1/3⇨x=5/6 & y=217/36, then eqn.(y-y1 ) = m(x-x1) ( by point slope form) gives 12x+36y – 227=0.Question 18: For the curve y = 4x3 − 2x5, find all the points at whichthe tangents passes through the origin.
- 68. Answer 18: Let (h,k) be the point on curve then k = 4h3 – 2h5 .......(i)Dy/dx = 12h2 – 10h4 at (h,k), eqn. Of tangent at (h,k) isY – k = (4h3 – 2h5 ) (x-h), it passes through origin⇨ k=12h3-10h5 , solvewith (i) ⇨ h=0 or ∓1 ⇨ k=0,-2,2Question 21: Find the equation of the normal to the curve y = x3 + 2x + 6which are parallel to the line x + 14y + 4 = 0.Answer 21: dy/dx= 3x2+2 , dy/dx=m(of line)= -1/14 , but normal to thecurve parallel to line ⇨ (-dx/dy)(x1,y1) = -1/14⇨ x1 = ∓2 & y1 = -6,18Eqns. Of normal are x+14y-254=0 & x+14y+86=0.Question 23: Prove that the curves x = y2 and xy = k cut at right angles if8k2 = 1. [Hint: Two curves intersect at right angle if the tangents to thecurves at the point of intersection are perpendicular to each other.]Answer: 23 after solving two curves , we get y=k1/3 & x=k2/3 , find slopes At this pointQuestion 25: Find the equation of the tangent to the curvewhich is parallel to the line 4x − 2y + 5 = 0.Answer 25: dy/dx= at (x1,y1) , slope of line is 2 by equating , weget x1 =41/48 & y1 = ¾ , eqn. Is 48x-24y=23.Ex. 6.4 Question 1:1. Using differentials, find the approximate value of each of the followingup to 3 places of decimal(i) (ii) (iii) (iv) (v) (vi)(vii) (viii) (ix) (x) (xi)(xii) (xiii) (xiv) (xv)Answer(i) let x=25 , △x=0.3 (iii) let x= 0.64, △x= -0.04 , = 0.8 +△y& △y = (dy/dx).△x= -(0.04)/2x0.8= -0.025∴ = .775
- 69. (iv) let x= 0.008, dx=0.001 , △y= (dy/dx) x [ △y/△x=dy/dx (app.)]△y = .001= .001=1/120=.008(x+△x)1/3 – x1/3 = (.009)1/3 – 0.2⇨ (.009)1/3 = 0.2+△y= 0.2+.008=.208(app.)(v) let x= 1, △x= -0.001 , dy △y = -.0001 ,answer is (0.999)1/10=0 .9999(viii) let x= 256 △x=-1 , △y = (x+△x)1/4 – x1/4 ⇨(255)1/4 = 4+△y ,dy/dx=1/(4x3/4) ∴ △y= (dy/dx).△x= . (-1)= -1/256 =-0.0039,so (255)1/4 = 4+△y = 3.9961.(xii) let x= 27, △x= -0.43 ,△y= . (-0.43)= -0.015926 , ans. =2.984(app)(xiv) let x= 4, △x= -0.032 , △y= (3√4/2)x-0.032=0.096 ,ans. =8-.096=7.904Question 5: Find the approximate change in the surface area of a cube ofside x metres caused by decreasing the side by 1%Answer 5: approximate change in the surface area=△s= ds/dx.△x= 12x. (-0.01x)= -0.12x2 m2Question 7: If the radius of a sphere is measured as 9 m with an error of0.03 m, then find the approximate error in calculating in surface area.Answer 7: △s = ds/dr.△r= 8пr .(0.03)= 2.16п m2EX. 6.5 Question 2: Find the maximum and minimum values, if any,of the following functions given by(i) f(x) = |x + 2| − 1 (ii) g(x) = − |x + 1| + 3 (iii) h(x) = sin(2x) + 5(iv) f(x) = |sin 4x + 3| (v) h(x) = x + 1, x (−1, 1)Answer: (i) |x + 2|≥0, x ∊R ∴ f(x)≥-1(mini.), no max. Value same as (ii)Part (iii) -1≤sin2x≤1, so max. Valueof f(x) is 6 & mini. Value is 4 (iv) max.Value of f(x) is |1+3|=4 & mini. Value is 2. (v) no max. Or mini. However,greatest & least value as at x=1,-1 f(x)=2 & 0.
- 70. Question 3: Find the local maximum value and the local minimum value ofthe functions(if any) in the given intervals: (iii) h(x) = sinx + cosx,(0, п/2) (vi) g(x) = x/2+2/x, x>0 (Vi)f(x)=x , x>0Answer: 3 (iii) h’(x)=cosx(1-tanx)=0⇨ tanx=1,x=п/4∊(0, п/2)By 1st derivative test, at x= п/4, we will check nature of h’(x)X<п/4 to x>п/4 , h’(x) changes it’s sign from +ve to –ve , local max.Value at п/4 is f(п/4) = √2 or second derivative test , we will findh’’(x)= -sinx – cosx, put x=п/4 ,we get h’’(x) <0⇨ max. At п/4.(vi) g’(x)= ½ - 2/x2 =0⇨ x=∓2, x=-2 is rejected so we will check at x=2 Both tests can be applied g,(x) changes its sign from –ve to +ve as xincreases , g(x) has local minima at x=2 & local mini. Value is 2 0r secondderivative test ⇨ g’’(x) = 4x-3 >0 at x=2.(vi) f’(x) =0 ⇨ x=2/3 , f’(x) changes sign from +ve to –ve as xincreases through x=2/3, local max. Value is 2√3/9.Question 8: At what points in the interval [0, 2π+, does the function sin 2x attain its maximum value?Answer 8: f’(x)=2cos2x=0 ⇨ x= п/4, 3п/4 ,5п/4 ,7п/4 , so max. Value Of f(x) is 1 at x= п/4 ,5п/4.Question 11: It is given that at x = 1, the function x4− 62x2 + ax + 9 attainsits maximum value, on the interval [0, 2]. Find the value of a.Answer 11: f’(x) =4x3-124x+a =0 ⇨ x=1 , f’(1) =0 ⇨ a=120, f’’(x)<0 at 1 F(x) has max. At x=1, when a=120Question 12: Find the maximum and minimum values of x + sin 2xon [0, 2π+.Answer 12: f’(x) = 1+2cos2x=0 ⇨ cos2x= -1/2⇨ 2x= 2п/3, 4п/3 ,8п/3,10п/3 ⇨ x= п/3, 2п/3 ,4п/3, 5п/3 , we get max. F(x)=2п & mini. =0Question 17: A square piece of tin of side 18 cm is to made into a boxwithout top, by cutting a square from each corner and folding up theflaps to form the box. What should be the side of the square to becut off so that the volume of the box is the maximum possible?Answer 17: volume of box= x(18-2x)2 ( x is the side of square) Dv/dx= (18-2x)(18-6x)=0⇨ x=3, 9,x≠9, d2v/dx2 = -72< 0 at x=3*Question 18:A rectangular sheet of tin 45 cm by 24 cm is to be made
- 71. into a box without top, by cutting off square from each corner andfolding up the flaps. What should be the side of the square to be cutoff so that the volume of the box is the maximum possible?Answer 18: v= 2x(45-2x)(12-x), dv/dx= 12(x2-23x+90) =0 ⇨ x= 5,18 d2v/dx2 < 0 at x=5 (x can’t be greater than 12)Question 20 Show that the right circular cylinder of the given surface andmax. Volume is such that its height is equal to the diameter of the base.Answer 20: S= 2пr2 +2пrh ⇨ h = , V =пr2( ) , dV/dr=0⇨S=6пr2 ⇨ h=2r, find d2v/dr2<0*Question 21: Of all the closed cylinder cans (right cylinder) of a gvenVolume of 100 cubic cm., find the dims. Of can which has the mini.Surface area?Answer 21: V= пr2h =100⇨ h= 100/ пr2 , S= 2пr2 +2пr(100/ пr2)Ds/dr=0⇨ пr3 =50 ⇨ r= (50/п)1/3 , find its 2nd derivative >0Question 22: A wire of length 28 m is to be cut into two pieces. Oneof the pieces is to be made into a square and the other into a circle.What should be the length of the two pieces so that the combinedarea of the square and the circle is minimum?Answer 22: let one part be x= 2пr & another 28-x be the per. Of square∴ r=x/2п & side of square is (28-x)/4 ⇨ A = x2/4п + [(28-x)/4]2dA/dx=0 ⇨ x=28п /(4+п) , find d2A/dx2 >0 ⇨ A is mini.Question 23:Prove that the volume of the largest cone that can beinscribed in a sphere of radius R is of the volume of the sphere.
- 72. Answer 23: Draw cone inside the circle R=OA=OP be radius & Ѳ=angleAOC (PC is height of cone),Consider △AOC, find V= 1/3 п (AC)2(PC)= 1/3 ПR3 sin2Ѳ(1+cosѲ) * ∵PC=PO+OC] , Dv/dѲ =0 ⇨ cosѲ=1/3 or -1cosѲ=1/3 [∵cosѲ≠-1 ∵ Ѳ ≠п], sinѲ = 2√2/3, find 2nd derivative <0which gives max. Volume ,put the values of sinѲ, cosѲ,we get V= 8/27(4/3ПR3)= 8/27(volume of sphere)Question 24: Show that the right circular cone of least curved surfaceand given volume has an altitude equal to time the radius of the base. [Answer 24: V= 1/3пr h....(i) ⇨ h=3V/ пr2 put it in S = пrl ⇨ 2S2 = п2r 2(r2+h2) ⇨ S2 = п2r 2(r2+(3V/ пr2)2)= P(let) , dP/dr =0 ⇨ r6 = 9V2/2п2 ......(ii) , 2nd der. >0 ∀ +ve r , from (i) & (ii)(by equating the values of V2), we get h=√2r.]Question 25: Show that the semi-vertical angle of the cone of themaximum volume and of given slant height is .Answer 25: Let l , r, h be slant height, radius, height & Ѳ be the semi-vertical angle , V= 1/3 п(lsinѲ)2(lcosѲ), dv/dѲ=0 ⇨ tanѲ =√2, find 2nd der. <0 ⇨ max. Volume at Ѳ= tan-1√2.Question 26: Show that the semi-vertical angle of the cone of givensurface and max. Volume is sin-1(1/3).Answer 26: S=пrl+пr2 ......(i) ⇨ l = (S-пr2)/пr put in volume=1/3пr²Take squaring let it P, then derivative =0⇨ S=4пr², 2nd der. <0, put theValue of S in (i) ⇨ r/l= sinѲ=1/3Question 28:For all real values of x, the minimum value of is(A) 0 (B) 1 (C) 3 (D) [ ans. (D), dy/dx=0⇨x=1,-1 & mini. At 1]
- 73. Question 29:The maximum value of is(A) (B) (C) 1 (D) 0 [ans. (C) dy/dx=0⇨x=1/2]Misc. Question 1: Using differentials, find the approximate value ofeach of the following. (a) (b)Answer 1: (a) (17)1/4/3, let x=16 & △x= 1, △y=(dy/dx).△x=1/32∴ (17)1/4 = 2.03125 & ans. is 2.03125/3=0.677(app.) (b) x=32, △x=1(33)1/5 =2+(1/80)=161/80 ∴ (1/33)1/5 = 80/161=0.497(app.)Question 2:Show that the function given by has maximum at x = e.Answer 2: f’(x)= (1-logx)/x2 =0 ⇨ x=e,show f’’(x)<0 at x=eQuestion 3:The two equal sides of an isosceles triangle with fixedbase b are decreasing at the rate of 3 cm per second. How fast is thearea decreasing when the two equal sides are equal to the base?Answer 3: let AB=AC=x & BC=b, L is mid point of BC.Consider △ABLAL= , Area = ½ . b. , dA/dt= (dA/dx).(dx/dt) at x=bWhere dx/dt =3cm./sec (given). Ans. is √3 bcm2 /sec.Question 6: Find the intervals in which the function f given by is (i) increasing (ii) decreasingAnswer 6: f’(x)= , 4-cosx>0 ∵ -1≤cosx≤1 ∴ f’(x)>0or <0
- 74. Depend on cosx>0 or<0 ∴ f is ↑ when 0<x<п/2 , 3п/2<x<2п & ↓ whenп/2<x<3п/2Question 7:Find the intervals in which the function f given by is (i) increasing (ii) decreasingAnswer 7: f’(x) 3x – 3/x4 >0 ⇨ (x3-1)(x3+1)>0 ⇨ (x3-1)>0 &(x3+1)>0 2 Or (x3-1)<0 & (x3+1)<0⇨ f is ↑ when x<-1 & x>1, ↓ when -1<x<1Question 8:Find the maximum area of an isosceles triangle inscribedin the ellipse with its vertex at one end of the major axis.Answer 8: let end vertex be A(a,0), P(acosѲ, bsinѲ)& Q be two on ellipsePQ intersect x-axis of ellipse at R. Area of isosceles △APQ = ½.PQ.AM=1/2 (2bsinѲ)(a-acosѲ) = ab(sinѲ-(1/2) sin2Ѳ) , dA/dѲ=0⇨ Ѳ=2п/3Show 2nd der. <0 at Ѳ=2п/3, max. Area is (3√3/4)ab at Ѳ=2п/3*Question 9:A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is8 m3. If building of tank costs Rs 70 per sq meters for the base and Rs45 per square metre for sides. What is the cost of least expensive tank?Answer 9: V= 2xy =8 ⇨ xy=4...(i) Area of base = xy, Area of sides =2.2(x+y)=4(x+y), cost of construction = Rs. {70xy+180(x+y)}.....(ii)From (i) & (ii) cost = 280+180(x+4/x), dC/dx =0 ⇨ x=∓2⇨ Cost is mini. At x=2∴ y=2 , least cost is 1000.*Question 10: The sum of per. Of a circle and a square is k, where k issome constant. Prove that the sum of their area is least when the sideof square is double the radius of the circle.Answer 10: 4x+2пr =k (x is a side of square, r is radius) ⇨ x = (k-2пr)/4 Put the value of x in A= x2+пr2 , dA/dr =0 ⇨ r = k/(8+2п), show 2nd der.>0 at r, then put the value of r in x ⇨ x=2r*Question 11:A window is in the form of rectangle surmounted by asemicircular opening. The total perimeter of the window is 10 m. Find thedimensions of the window to admit maximum light through the wholeopening.Answer 11: let width, length be x, 2r so per. = 2x+2r+1/2 .2пr= 10...(i)A = 2RX+1/2 Пr2 ...(ii) , from (i) & (ii) ⇨ A= 10r – (1/2п+2)r2 , dA/dr=0
- 75. ⇨r = , find 2nd der.<0 at r , length=2r=20/(п+4), breadth=*Question 12:A point on the hypotenuse of a triangle is at distance aand b from the sides of the triangle.Show that the minimum lengthof the hypotenuse isAnswer 12: Draw a right angled △, AB, BC, AC are per. , base & hyp. Resp.Draw PL parallel to BC, PM is parallel to AB . angle C is Ѳ. Consider △APL & PCM ⇨ l= AP+PC ⇨asecѲ+bcosecѲ, 0<Ѳ<п/2, dl/dѲ=0⇨tanѲ=(b/a)1/3 , show 2nd der. >0, least value of l=Question 13: Find the points at which the function f given byHas (i) local maxima (ii) local minima (iii) point of inflexionAnswer 13: f’(x)=0⇨ x=2, -1, 2/7 f(x) is mini. At x=2, x=-1 is point ofinflexion( does not change its sign) & max. At x=2/7Question 15: Show that the altitude of the right circular cone ofmaximum volume that can be inscribed in a sphere of radius r is .Answer 15: Take OB is radius, AM is height of cone, BC is base of coneM is mid point of BC, angle BOM=Ѳ. AM= AO+OM=r(1+cosѲ)V= 1/3 пr3sin2Ѳ(1+cosѲ, dV/dѲ= 0⇨ cosѲ=1/3,cosѲ ≠-1,show 2nd der.<0, then find altitude = r(1+cosѲ)= 4r/3Question 17: Show that the height of the cylinder of maximum volumethat can be inscribed in a sphere of radius R is . Also find themaximum volume.Answer17 : Radius of sphere is R, x & h diameter , height of cylinder h2+x2=(2R)2 ....(i) , V= п.(x/2)2.h = ¼ пh(4R2-h2) from(i), dV/dh=0⇨h = , show 2nd der. <0 at h & max. Volume is .Question 18: Show that height of the cylinder of greatest volume
- 76. which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is . Answer 18: vertical angle be (cone) OO’ is height of cylinder = VO-VO’ = h-xcot ( O’ ,O are centre of top & bottom of cly.), V= Пx2( h-xcot ) Dv/dx=0 ⇨ x=(2п/3)tan , show 2nd der. <0 & max. Volume at x. Question 20:The slope of the tangent to the curve at the point (2, −1) is(A) (B) (C) (D) Answer 20: (B) , put x=2 ⇨ t = - 5,2 then put t=2 ⇨ y=-1 dy/dx= 4t-2/2t+3 = 6/7 at t=2. ASSIGMENT OF DIFFERENTIATION Question 1Show that y = aex and y = be –x cut at right angles if ab=1 [by equating , we get ex = ⇨ x= ½ log ( b/a) , find slopes(dy/dx) at pt. of intersection is (½ log ( b/a , ). Question 2 (i) If y +x = 1, prove that dy/dx= (-1)[Hint: put y=sinѲ & x= sin , use formula of sin((ii) If cos-1 ) = tan-1a , find dy/dx.[let cos(tan-1a )= k(constant), then assume c= 1-k/1+k , dy/dx= y/x](iii) If = , prove that dy/dx =(iv) If xm.yn = (x+y)m+n, then find dy/dx. [ y/x]Question 3 Differentiate w.r.t. x :**(i) Using logarithmic differentiation, differentiate:
- 77. Solution: x logx(ii) + (iii) (iogx) + xQuestion 4 (i) If = , prove that dy/dx =(ii) If f(1)= 4,f’(1)=2,find d/dx{logf(ex)} at the point x =0.[1/2](iii)If y = ,show that (2y – 1)dy/dx =1. a (t+1/t)(iv) If x = (t+1/t) , y= a where a>0,a≠1,t≠0, find dy/dx.[Hint: take dy/dt & dx/dt , then find dy/dx = ylogy/ax. ]Question5(i)differentiate: Sec-1(1/(2x2 – 1)),w.r.t.sin-1(3x –4x3).[Hint: let u=1st fn. & v= 2nd fn. , find du/dv = 1](ii)differentiate: tan-1 ( ),w.r.t. sin-1 ( ) if 1<x<1;x≠0[ du/dv= ¼, put x=tanѲ⇨ u=Ѳ/2, v=2Ѳ , u&v as assumed above] (msin-1x)(iii) If y = e , show that (1-x2)y2 – xy1 – m2y= 0.Question 6 Water is driping out from a conical funnel, at theuniform rate of 2cm3/sec. through a tiny hole at the vertex atthe bottom. When the slant height of the water is 4cm.,findthe rate of decrease of the slant height of the water given that the vertical angle of the funnel is 1200 .[Let l is slant height ,V = 1/3. .l /2.l/2= l3/8(vertical angle will be 600 (half cone), take dv/dt=-2cm3/sec. ⇨l=-1/3 cm/s.][Hint: put y=sinѲ & x= sin , use formula of sin(
- 78. (ii) If cos-1 ) = tan-1a , find dy/dx.[let cos(tan-1a )= k(constant), then assume c= 1-k/1+k , dy/dx= y/x](iii) If = , prove that dy/dx =(iv) If xm.yn = (x+y)m+n, then find dy/dx. [ y/x]Question 3 Differentiate w.r.t. x :**(i) Using logarithmic differentiation, differentiate:Solution: x logx(ii) + (iii) (iogx) + xQuestion 4 (i) If = , prove that dy/dx =(ii) If f(1)= 4,f’(1)=2,find d/dx{logf(ex)} at the point x =0.[1/2](iii)If y = ,show that (2y – 1)dy/dx =1. a (t+1/t)(iv) If x = (t+1/t) , y= a where a>0,a≠1,t≠0, find dy/dx.[Hint: take dy/dt & dx/dt , then find dy/dx = ylogy/ax. ]Question5(i)differentiate: Sec-1(1/(2x2 – 1)),w.r.t.sin-1(3x –4x3).[Hint: let u=1st fn. & v= 2nd fn. , find du/dv = 1](ii)differentiate: tan-1 ( ),w.r.t. sin-1 ( ) if 1<x<1;x≠0[ du/dv= ¼, put x=tanѲ⇨ u=Ѳ/2, v=2Ѳ , u&v as assumed above] (msin-1x)(iii) If y = e , show that (1-x2)y2 – xy1 – m2y= 0.Question 6 Water is driping out from a conical funnel, at the uniform
- 79. rate of 2cm3/sec. through a tiny hole at the vertex at the bottom.When the slant height of the water is 4cm.,find the rate of decreaseof the slant height of the water given that the vertical angle of the funnel is 1200 .[Hint: Let l is slant height ,V = 1/3. .l /2.l/2= l3/8(vertical anglewill be 600 (half cone), take dv/dt=-2cm3/sec. ⇨l=-1/3 cm/s.]**Question 7(i) Let f be differentiable for all x. If f(1)=-2 and f f f `(x) ≥2 ∀ x∊[1, 6], then prove f(6) ≥8.[ use L.M.V.Thm.,f`(c)≥2,c∊[1, 6]](ii) If the function f(x)= x3 – 6x2+ax+b defined on [1, 3] satisfies therolle’s theorem for c = (2 +i)/ , then p.t. a = 11 & b∊R.[Hint: Take f(1)=f(3) , use rolle’s thm. f`(c)=0⇨ a=11]Question 8 (i) Show that f(x)= x/sinx is increasing in (0, п/2)*HINT: f’(x)>0 , tanx >x+(ii) Find the intervals of increase and decrease for f(x) = x3 + 2x2 – 1.[Answer is increasing in (-∞, -4/3)U(0, ∞) & decreasing in (-4/3, 0)](iii) Find the interval of increase&decrease for f(x) =log(1+x)-(x/1+x) ORProve that x/(1+x) < log(1+x) < x for x > 0.[ Hint: f(x)strictly ↑ in [0, ∞) , x>0 ⇨f(x)>f(0), let g(x)=x-log(1+x)g(x)>0 ↑ in [0,∞) & f(x) ↓ in (-∞, 0].](iv) For which value of a , f(x)=a(x+sinx)+a is increasing.[Hint: f’(x) a(1+cosx) ≥0 ⇨ a>0 ∵ -1≤cosx≤1]**Question 9 Problem: Using differentials, approximate the expression [ arctan(1.05) = tan-1(1.05)]Solution:We letHence, x = 0.05 and y = /4.Differentiating, we obtain
- 80. Substituting, we getQuestion 10 Find the stationary points of the function f(x) =3x4 – 8x3+6x2 and distinguish b/w them. Also find the local max.And local mini. Values, if they exist.* f’(x)=0⇨ x=0,1 f has local mini. At x=0∵f’’>0 & f’’’’(1)≠0, f has point ofinflexion at x=1,f(1)=1]Question 11 Show that the semi – vertical angle of right circular cone ofgiven total surface area and max. Volume is sin-1 1/3.[Hint: take S=Пr(l+r) ⇨ l= S/пr – r , take derivative of V² OR can usetrigonometric functions for l & h]Question 12 A window has the shape of a rectangle surmounted by anequilateral ∆. If the perimeter of the window is 12 m., find the dimensionsof the rectangle so that it may produce the largest area of the window.[Hint: let x=length, y=breadth, then y=6 – 3y/2, A= XY+ X2 /4, takederivative of A & it is max. ,x=4(6+ )/11 ,y=6(5 )/11]

No public clipboards found for this slide

Be the first to comment