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Lesson 11: Implicit Differentiation (Section 21 slides)
 

Lesson 11: Implicit Differentiation (Section 21 slides)

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Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of ...

Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.

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    Lesson 11: Implicit Differentiation (Section 21 slides) Lesson 11: Implicit Differentiation (Section 21 slides) Presentation Transcript

    • Section 2.6 Implicit Differentiation V63.0121.021, Calculus I New York University October 12, 2010 Announcements Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2 Midterm next week. Covers §§1.1–2.5 . . . . . .
    • Announcements Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2 Midterm next week. Covers §§1.1–2.5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 2 / 34
    • Objectives Use implicit differentation to find the derivative of a function defined implicitly. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 3 / 34
    • Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 4 / 34
    • Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . x2 + y2 = 1 at the point (3/5, −4/5). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
    • Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . x2 + y2 = 1 at the point (3/5, −4/5). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
    • Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . x2 + y2 = 1 . at the point (3/5, −4/5). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
    • Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . x2 + y2 = 1 . at the point (3/5, −4/5). Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
    • Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . x2 + y2 = 1 . at the point (3/5, −4/5). Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) dy −2x x Differentiate: =− √ =√ dx 2 1−x 2 1 − x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
    • Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . x2 + y2 = 1 . at the point (3/5, −4/5). Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) dy −2x x Differentiate: =− √ =√ dx 2 1−x 2 1 − x2 dy 3/5 3/5 3 Evaluate: =√ = = . dx x=3/5 1 − (3/5)2 4/5 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
    • Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . x2 + y2 = 1 . at the point (3/5, −4/5). Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) dy −2x x Differentiate: =− √ =√ dx 2 1−x 2 1 − x2 dy 3/5 3/5 3 Evaluate: =√ = = . dx x=3/5 1 − (3/5)2 4/5 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 5 / 34
    • Motivating Example, another way We know that x2 + y2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
    • Motivating Example, another way We know that x2 + y2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 We could differentiate this equation to get 2x + 2f(x) · f′ (x) = 0 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
    • Motivating Example, another way We know that x2 + y2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 We could differentiate this equation to get 2x + 2f(x) · f′ (x) = 0 We could then solve to get x f′ (x) = − f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 6 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. . x . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. . x . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. . x . . l .ooks like a function . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the . curve resembles the graph of a function. . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the . curve resembles the graph of a function. . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the . curve resembles the graph of a function. l .ooks like a function . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. . . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. . . x . . does not look like a function, but that’s OK—there are only two points like this . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally”, . x . almost everywhere and is differentiable . l .ooks like a function . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Yes, we can! The beautiful fact (i.e., deep theorem) is that this works! y . “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally”, . x . almost everywhere and is differentiable The chain rule then applies . for this local choice. l .ooks like a function . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 7 / 34
    • Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
    • Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
    • Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx Remember y is assumed to be a function of x! . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
    • Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx Remember y is assumed to be a function of x! dy x Isolate: =− . dx y . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
    • Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx Remember y is assumed to be a function of x! dy x Isolate: =− . dx y dy 3/5 3 Evaluate: = = . dx ( 3 ,− 4 ) 4/5 4 5 5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 8 / 34
    • Summary If a relation is given between x and y which isn’t a function: “Most of the time”, i.e., “at y . most places” y can be . assumed to be a function of x we may differentiate the . x . relation as is dy Solving for does give the dx slope of the tangent line to the curve at a point on the curve. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 9 / 34
    • Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 10 / 34
    • Another Example Example Find y′ along the curve y3 + 4xy = x2 + 3. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
    • Another Example Example Find y′ along the curve y3 + 4xy = x2 + 3. Solution Implicitly differentiating, we have 3y2 y′ + 4(1 · y + x · y′ ) = 2x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
    • Another Example Example Find y′ along the curve y3 + 4xy = x2 + 3. Solution Implicitly differentiating, we have 3y2 y′ + 4(1 · y + x · y′ ) = 2x Solving for y′ gives 3y2 y′ + 4xy′ = 2x − 4y (3y2 + 4x)y′ = 2x − 4y 2x − 4y =⇒ y′ = 2 3y + 4x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 11 / 34
    • Yet Another Example Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 12 / 34
    • Yet Another Example Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). Solution Differentiating implicitly: 5y4 y′ + (2x)y3 + x2 (3y2 y′ ) = y′ sin(x2 ) + y cos(x2 )(2x) Collect all terms with y′ on one side and all terms without y′ on the other: 5y4 y′ + 3x2 y2 y′ − sin(x2 )y′ = −2xy3 + 2xy cos(x2 ) Now factor and divide: 2xy(cos x2 − y2 ) y′ = 5y4 + 3x2 y2 − sin x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 12 / 34
    • Finding tangent lines with implicit differentitiation . Example Find the equation of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6). . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
    • Finding tangent lines with implicit differentitiation . Example Find the equation of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6). . Solution dy dy 3x2 + 2x Differentiate: 2y = 3x2 + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 33 11 = =− =− . dx (3,−6) 2(−6) 12 4 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
    • Finding tangent lines with implicit differentitiation . Example Find the equation of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6). . Solution dy dy 3x2 + 2x Differentiate: 2y = 3x2 + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 33 11 = =− =− . dx (3,−6) 2(−6) 12 4 11 Thus the equation of the tangent line is y + 6 = − (x − 3). 4 . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 13 / 34
    • Recall: Line equation forms slope-intercept form y = mx + b where the slope is m and (0, b) is on the line. point-slope form y − y0 = m(x − x0 ) where the slope is m and (x0 , y0 ) is on the line. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 14 / 34
    • Horizontal Tangent Lines Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 15 / 34
    • Horizontal Tangent Lines Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We have to solve these two equations: . . 3x2 + 2x 2 3 2 = 0 1 y = x. + x . [(x, y) is on the curve] 2 . 2y [tangent line is horizontal] . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 15 / 34
    • Solution, continued Solving the second equation gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
    • Solution, continued Solving the second equation gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
    • Solution, continued Solving the second equation gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. Substituting x = 0 into the first equation gives y2 = 03 + 02 = 0 =⇒ y = 0 which we’ve disallowed. So no horizontal tangents down that road. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
    • Solution, continued Solving the second equation gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. Substituting x = 0 into the first equation gives y2 = 03 + 02 = 0 =⇒ y = 0 which we’ve disallowed. So no horizontal tangents down that road. Substituting x = −2/3 into the first equation gives ( ) ( ) 2 3 2 2 4 2 y = − 2 + − = =⇒ y = ± √ , 3 3 27 3 3 so there are two horizontal tangents. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 16 / 34
    • Horizontal Tangents ( ) . −2, 3 2 √ 3 3 . . . ( ) . −2, − 3 2 √ 3 3 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 17 / 34
    • Horizontal Tangents ( ) . −2, 3 2 √ 3 3 . . . ( ) . −2, − 3 2 √ n . ode 3 3 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 17 / 34
    • Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
    • Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx Tangent lines are vertical when = 0. dy . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
    • Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx Tangent lines are vertical when = 0. dy Differentiating x implicitly as a function of y gives dx dx dx 2y 2y = 3x2 + 2x , so = 2 (notice this is the dy dy dy 3x + 2x reciprocal of dy/dx). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
    • Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx Tangent lines are vertical when = 0. dy Differentiating x implicitly as a function of y gives dx dx dx 2y 2y = 3x2 + 2x , so = 2 (notice this is the dy dy dy 3x + 2x reciprocal of dy/dx). We must solve . . 2y y2 = x3 + x2 =0 3x2 + 2x 1 . [(x, y). is on the curve] 2 . [tangent line . . . is vertical] . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 18 / 34
    • Solution, continued Solving the second equation gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
    • Solution, continued Solving the second equation gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). Substituting y = 0 into the first equation gives 0 = x3 + x2 = x2 (x + 1) So x = 0 or x = −1. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
    • Solution, continued Solving the second equation gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). Substituting y = 0 into the first equation gives 0 = x3 + x2 = x2 (x + 1) So x = 0 or x = −1. x = 0 is not allowed by the first equation, but dx = 0, dy (−1,0) so here is a vertical tangent. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 19 / 34
    • Tangents ( ) . −2, 3 2 √ 3 3 . . −1, 0) . ( . . ( ) . −2, − 3 2 √ n . ode 3 3 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 20 / 34
    • Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 21 / 34
    • Orthogonal Families of Curves y . .xy = 1 xy = c x2 − y2 = k . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy = .xy 2 = 1 xy = c x2 − y2 = k . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . 1 − = .xy . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . 1 − 2 = − .xy = .xy . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = 1 xy = c x2 − y2 = k . x . .xy = −1 = −2 .xy = .xy 3 − . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . .xy = −1 = −2 x .xy = .xy 3 − . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . .xy = −1 = −2 x x .xy = .xy 3 − . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . .xy = −1 x2 = −2 x x .xy = .xy 3 − . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . .xy = −1 x2 = −2 x x . 2 − y2 = −1 .xy = x .xy 3 − . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . .xy = −1 x2 = −2 x x . 2 − y2 = −1 .xy = x .xy 3 . 2 − y2 = −2 − x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Orthogonal Families of Curves y . .xy .xy = 3 = .xy 2 = . 2 − y2 = 3 . − y2 = 2 . 2 − y2 = 1 1 xy = c x2 − y2 = k . x . .xy = −1 x2 = −2 x x . 2 − y2 = −1 .xy = x .xy 3 . 2 − y2 = −2 − x2 . − y2 = −3 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 22 / 34
    • Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution y In the first curve, y + xy′ = 0 =⇒ y′ = − x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
    • Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution y In the first curve, y + xy′ = 0 =⇒ y′ = − x x In the second curve, 2x − 2yy′ = 0 =⇒ y′ = y . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
    • Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution y In the first curve, y + xy′ = 0 =⇒ y′ = − x x In the second curve, 2x − 2yy′ = 0 =⇒ y′ = y The product is −1, so the tangent lines are perpendicular wherever they intersect. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 23 / 34
    • Music Selection “The Curse of Curves” by Cute is What We Aim For . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 24 / 34
    • Ideal gases The ideal gas law relates temperature, pressure, and volume of a gas: PV = nRT (R is a constant, n is the amount of gas in moles) . Ima . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 25 / 34
    • Compressibility Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 26 / 34
    • Compressibility Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. Approximately we have ∆V dV ∆V ≈ = −βV =⇒ ≈ −β∆P ∆P dP V The smaller the β, the “harder” the fluid. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 26 / 34
    • Compressibility of an ideal gas Example Find the isothermic compressibility of an ideal gas. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 27 / 34
    • Compressibility of an ideal gas Example Find the isothermic compressibility of an ideal gas. Solution If PV = k (n is constant for our purposes, T is constant because of the word isothermic, and R really is constant), then dP dV dV V ·V+P = 0 =⇒ =− dP dP dP P So 1 dV 1 β=− · = V dP P Compressibility and pressure are inversely related. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 27 / 34
    • Nonideal gasses Not that there's anything wrong with that Example The van der Waals equation H .. makes fewer simplifications: ( ) O . . xygen . . H n2 . P + a 2 (V − nb) = nRT, V H .. where P is the pressure, V the O . . xygen H . ydrogen bonds volume, T the temperature, n H .. . the number of moles of the gas, R a constant, a is a measure of O . . xygen . . H attraction between particles of the gas, and b a measure of H .. particle size. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 28 / 34
    • Nonideal gasses Not that there's anything wrong with that Example The van der Waals equation makes fewer simplifications: ( ) n2 P + a 2 (V − nb) = nRT, V . where P is the pressure, V the volume, T the temperature, n the number of moles of the gas, R a constant, a is a measure of attraction between particles of the gas, and b a measure of particle size. . . ikimedia Commons W . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 28 / 34
    • Compressibility of a van der Waals gas Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
    • Compressibility of a van der Waals gas Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
    • Compressibility of a van der Waals gas Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Question . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
    • Compressibility of a van der Waals gas Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Question What if a = b = 0? . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
    • Compressibility of a van der Waals gas Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Question What if a = b = 0? dβ Without taking the derivative, what is the sign of ? db . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
    • Compressibility of a van der Waals gas Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Question What if a = b = 0? dβ Without taking the derivative, what is the sign of ? db dβ Without taking the derivative, what is the sign of ? da . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 29 / 34
    • Nasty derivatives dβ (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 ) =− db (2abn3 − an2 V + PV3 )2 ( ) nV3 an2 + PV2 = −( )2 < 0 PV3 + an2 (2bn − V) dβ n2 (bn − V)(2bn − V)V2 =( )2 > 0 da 3 2 (2bn − V) PV + an (as long as V > 2nb, and it’s probably true that V ≫ 2nb). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 30 / 34
    • Outline The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 31 / 34
    • Using implicit differentiation to find derivatives Example dy √ Find if y = x. dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 32 / 34
    • Using implicit differentiation to find derivatives Example dy √ Find if y = x. dx Solution √ If y = x, then y2 = x, so dy dy 1 1 2y = 1 =⇒ = = √ . dx dx 2y 2 x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 32 / 34
    • The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
    • The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
    • The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp Now, differentiate implicitly: dy dy p xp−1 qyq−1 = pxp−1 =⇒ = · q−1 dx dx q y . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
    • The power rule for rational powers Theorem p p/q−1 If y = xp/q , where p and q are integers, then y′ = x . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp Now, differentiate implicitly: dy dy p xp−1 qyq−1 = pxp−1 =⇒ = · q−1 dx dx q y Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so xp−1 xp−1 = p−p/q = xp−1−(p−p/q) = xp/q−1 . . . . . . V63.0121.021, Calculus I yq−1 (NYU) x Section 2.6 Implicit Differentiation October 12, 2010 33 / 34
    • Summary Implicit Differentiation allows us to pretend that a relation describes a function, since it does, locally, “almost everywhere.” The Power Rule was established for powers which are rational numbers. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 12, 2010 34 / 34