Rate of change and tangent lines

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find average and instantaneous velocity?,find the tangent line? normal line?

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  • For analytic techniques you will need a good algebraic background.
  • Transcript of "Rate of change and tangent lines"

    1. 1. 2.4 Rates of Change and Tangent Lines Devil’s Tower, Wyoming
    2. 2. 2 1 2 1 Average rate of change change in y y y y change in x x x x − ∆ = = = − ∆ If f(t) represents the position of an object as a function of time, then the rate of change is the velocity of the object. Average rate of change (from bc) Average rate of change of f(x) over the interval [a,b] ab afbf − − = )()(
    3. 3. Find the average rate of change of f (t) = 2 + cost on [0, π] F(b)= f(π) = 2 + cos (π) = 2 – 1 = 1 F(a)= f(0) = 2 + cos (0) = 2 + 1 = 3 1. Calculate the function value (position) at each endpoint of the interval The average velocity on [0, π] is 0.63366 2. Use the slope formula 63366.0 2 0 31)()( =≈ − = − − = − − = ππab afbf
    4. 4. Consider a graph of displacement (distance traveled) vs. time. time (hours) distance (miles) Average velocity can be found by taking: change in position change in time s t ∆ = ∆ t∆ s∆ A B ( ) ( ) ave f t t f ts V t t + ∆ −∆ = = ∆ ∆ The speedometer in your car does not measure average velocity, but instantaneous velocity. (The velocity at one moment in time.) The velocity problem
    5. 5. The slope of a line is given by: y m x ∆ = ∆ x∆ y∆ The slope of a curve at (1,1) can be approximated by the slope of the secant line through (1,1) and (4,16). 5= We could get a better approximation if we move the point closer to (1,1). ie: (3,9) y x ∆ ∆ 9 1 3 1 − = − 8 2 = 4= Even better would be the point (2,4). y x ∆ ∆ 4 1 2 1 − = − 3 1 = 3= → y x ∆ ∆ 16 1 4 1 − = − 15 3 =
    6. 6. The slope of a line is given by: y m x ∆ = ∆ x∆ y∆ If we got really close to (1,1), say (1.1,1.21), the approximation would get better still y x ∆ ∆ 1.21 1 1.1 1 − = − .21 .1 = 2.1= How far can we go? ( ) 2 f x x= →
    7. 7. If we try to apply the same formula to find The instantaneous velocity and evaluate the velocity at an instant (a,f(a)) not an interval , we will find it 0 0)()( t = − − = ∧ ∧ aa afafs Which is undefined, so the best is to make Δx as small as experimentally possible Δx 0 ax afxf ax − − → )()( lim The instantaneous velocity at the point (a,f(a)) = ( ) ( ) ( ) 0 lim t f t t f tds V t dt t∆ → + ∆ − = = ∆
    8. 8. The slope of the secant line= the average rate of change = The average velocity The slope of the curve at a point = the slope of the tangent line of the curve at this point = instantaneous velocity = 2 1 2 1 Average rate of change change in y y y y change in x x x x − ∆ = = = − ∆ h xfhxf h )()( lim0 −+ → = − − → ax afxf ax )()( lim
    9. 9. Other form for Slope of secant line of tangent line sec ( ) ( )y f a h f a m x h ∆ + − = = ∆ tan 0 ( ) ( ) limh f a h f a m h → + − = Let h = x - a Then x = a + h
    10. 10. Rates of Change: Average rate of change = ( ) ( )f x h f x h + − Instantaneous rate of change = ( ) ( ) ( ) 0 lim h f x h f x f x h→ + − ′ = These definitions are true for any function. ( x does not have to represent time. )
    11. 11. Analytic Techniques Rewrite algebraically if direct substitution produces an indeterminate form such as 0/0 • Factor and reduce • Rationalize a numerator or denominator • Simplify a complex fraction When you rewrite you are often producing another function that agrees with the original in all but one point. When this happens the limits at that point are equal.
    12. 12. Find the indicated limit 2 3 6 lim 3x x x x→− + − + 3 lim ( 2) x x →− − 3 ( 3)( 2) lim 3x x x x→− + − + = - 5 direct substitution fails Rewrite and cancel now use direct sub. 0 0
    13. 13. ( )1f 1 1 h+ ( )1f h+ h slope y x ∆ = ∆ ( ) ( )1 1f h f h + − = slope at ( )1,1 ( ) 2 0 1 1 lim h h h→ + − = 2 0 1 2 1 lim h h h h→ + + − = ( ) 0 2 lim h h h h→ + = 2= The slope of the curve at the point is:( )y f x= ( )( ),P a f a ( ) ( ) 0 lim h f a h f a m h→ + − = →
    14. 14. The slope of the curve at the point is:( )y f x= ( )( ),P a f a ( ) ( ) 0 lim h f a h f a m h→ + − = ( ) ( )f a h f a h + − is called the difference quotient of f at a. If you are asked to find the slope using the definition or using the difference quotient, this is the technique you will use. →
    15. 15. In the previous example, the tangent line could be found using .( )1 1y y m x x− = − The slope of a curve at a point is the same as the slope of the tangent line at that point. If you want the normal line, use the negative reciprocal of the slope. (in this case, ) 1 2 − (The normal line is perpendicular.) →
    16. 16. Example 4: a Find the slope at .x a= ( ) ( ) 0 lim h f a h f a m h→ + − = 0 1 1 lim h a h a h→ − += ( ) ( )0 1 lim hh a a h a a h→ − + = × + ( )0 lim h a a h h a a h→ − − = × + 2 1 a = − Let ( ) 1 f x x = On the TI-89: limit ((1/(a + h) – 1/ a) / h, h, 0) F3 Calc Note: If it says “Find the limit” on a test, you must show your work! → ( )a a h+ ( )a a h+ ( )a a h+ 0
    17. 17. Example 4: b Where is the slope ? 1 4 − Let ( ) 1 f x x = 2 1 1 4 a − = − 2 4a = 2a = ± On the TI-89: Y= y = 1 / x WINDOW 6 6 3 3 scl 1 scl 1 x y x y − < < − < < = = GRAPH →
    18. 18. Example 4: b Where is the slope ? 1 4 − Let ( ) 1 f x x = On the TI-89: Y= y = 1 / x WINDOW 6 6 3 3 scl 1 scl 1 x y x y − < < − < < = = GRAPH We can let the calculator plot the tangent: F5 Math A: Tangent ENTER 2 ENTER Repeat for x = -2 tangent equation →
    19. 19. Review: average slope: y m x ∆ = ∆ slope at a point: ( ) ( ) 0 lim h f a h f a m h→ + − = average velocity: ave total distance total time V = instantaneous velocity: ( ) ( ) 0 lim h f t h f t V h→ + − = If is the position function:( )f t These are often mixed up by Calculus students! So are these! velocity = slope π
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