Lesson 11: Implicit Differentiation

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Published on October 8, 2009

Author: leingang

Source: slideshare.net

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Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.

Section 2.6 Implicit Differentiation V63.0121.034, Calculus I October 7, 2009 Announcements Midterm next , covering §§1.1–2.4. . . Image credit: Telstar Logistics . . . . . .

Outline The big idea, by example Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .

Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . .

Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . .

Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . . .

Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) . . . . . .

Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =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 . . . . . .

Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =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 . . . . . .

Motivating Example y . Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =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 . . . . . .

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) . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally” and is . x . differentiable The chain rule then applies for this local . choice. . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally” and is . x . differentiable The chain rule then applies for this local . choice. . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally” and is . x . differentiable The chain rule then applies for this local . choice. l .ooks like a function . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the . graph of a function. So f(x) is defined “locally” and is . x . differentiable The chain rule then applies for this local choice. . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the . graph of a function. So f(x) is defined “locally” and is . x . differentiable The chain rule then applies for this local choice. . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the . graph of a function. l .ooks like a function So f(x) is defined “locally” and is . x . differentiable The chain rule then applies for this local choice. . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally” and is . . x . differentiable The chain rule then applies for this local choice. . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally” and is . . x . differentiable The chain rule then applies for this local choice. . . . . . .

The beautiful fact (i.e., deep theorem) is that this works! “Near” most points on y . the curve x2 + y2 = 1, the curve resembles the graph of a function. So f(x) is defined “locally” and is . . x . differentiable . The chain rule then does not look like a applies for this local function, but that’s choice. OK—there are only two points like this . . . . . .

Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution (Implicit, with Leibniz notation) Differentiate. Remember y is assumed to be a function of x: dy 2x + 2y = 0, dx dy Isolate : dx dy x =− . dx y Evaluate: dy 3 /5 3 = = . dx ( 3 ,− 4 ) 4/5 4 5 5 . . . . . .

Summary If a relation is given between x y . and y, “Most of the time”, i.e., “at most places” y can be . assumed to be a function of x . we may differentiate the relation as is dy Solving for does give the dx slope of the tangent line to the curve at a point on the curve. . . . . . .

Mnemonic Explicit Implicit y = f(x) F(x, y) = k . . . . . .

Outline The big idea, by example Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .

Example Find the equation of the line tangent to the curve . y 2 = x 2 (x + 1 ) = x 3 + x 2 at the point (3, −6). . . . . . . .

Example Find the equation of the line tangent to the curve . y 2 = x 2 (x + 1 ) = x 3 + x 2 at the point (3, −6). . Solution Differentiating the expression implicitly with respect to x gives dy dy 3x2 + 2x 2y = 3x2 + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 11 = =− . dx (3,−6) 2(−6) 4 . . . . . .

Example Find the equation of the line tangent to the curve . y 2 = x 2 (x + 1 ) = x 3 + x 2 at the point (3, −6). . Solution Differentiating the expression implicitly with respect to x gives dy dy 3x2 + 2x 2y = 3x2 + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 11 = =− . dx (3,−6) 2(−6) 4 11 Thus the equation of the tangent line is y + 6 = − (x − 3). 4 . . . . . .

Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 . . . . . .

Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution . . . . . .

Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We solve for dy/dx = 0: 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y . . . . . .

Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We solve for dy/dx = 0: 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y The possible solution x = 0 leads to y = 0, which is not a smooth point of the function (the denominator in dy/dx becomes 0). . . . . . .

Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We solve for dy/dx = 0: 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y The possible solution x = 0 leads to y = 0, which is not a smooth point of the function (the denominator in dy/dx becomes 0). The possible solution x = − 2 yields y = ± 3√3 . 3 2 . . . . . .

Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 . . . . . .

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 2y = 3x2 + 2x , so dy dy dx 2y = 2 dy 3x + 2x . . . . . .

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 2y = 3x2 + 2x , so dy dy dx 2y = 2 dy 3x + 2x This is 0 only when y = 0. . . . . . .

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 2y = 3x2 + 2x , so dy dy dx 2y = 2 dy 3x + 2x This is 0 only when y = 0. We get the false solution x = 0 and the real solution x = −1. . . . . . .

Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). . . . . . .

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 + 2x cos(x2 ) Now factor and divide: 2x(y3 + cos x2 ) y′ = 5y4 + 3x2 y2 . . . . . .

Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. . . . . . .

Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution In the first curve, y y + xy′ = 0 =⇒ y′ = − x . . . . . .

Examples Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution In the first curve, y y + xy′ = 0 =⇒ y′ = − x In the second curve, x 2x − 2yy′ = 0 = =⇒ y′ = y The product is −1, so the tangent lines are perpendicular wherever they intersect. . . . . . .

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) . . Image credit: Scott Beale / Laughing Squid . . . . . .

. Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. . Image credit: Neil Better . . . . . .

. Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. . The smaller the β , the “harder” the fluid. Image credit: Neil Better . . . . . .

Example Find the isothermic 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. . . . . . .

Nonideal gasses Not that there’s anything wrong with that Example The van der Waals equation makes fewer simplifications: H .. ( ) O . . xygen . . n2 H 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 O . . xygen . . H measure of attraction between particles of the gas, H .. and b a measure of particle size. . . . . . .

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. . Image credit: Wikimedia Commons . . . . . .

Let’s find the 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 . . . . . .

Let’s find the 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 . . . . . .

Let’s find the 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 What if a = b = 0? . . . . . .

Let’s find the 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 What if a = b = 0? dβ Without taking the derivative, what is the sign of ? db . . . . . .

Let’s find the 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 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 . . . . . .

Nasty derivatives dβ (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 ) =− db (2abn3 − an2 V + PV3 )2 ( 2 ) nV3 an + PV2 = −( )2 < 0 PV3 + an2 (2bn − V) dβ n2 (bn − V)(2bn − V)V2 = ( )2 > 0 da PV3 + an2 (2bn − V) (as long as V > 2nb, and it’s probably true that V ≫ 2nb). . . . . . .

Outline The big idea, by example Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers . . . . . .

Using implicit differentiation to find derivatives Example dy √ Find if y = x. dx . . . . . .

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 . . . . . .

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 . . . . . .

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. We have dy dy p x p −1 yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1 dx dx q y . . . . . .

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. We have dy dy p x p −1 yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1 dx dx q y Now yq−1 = x(p/q)(q−1) = xp−p/q so x p −1 = xp−1−(p−p/q) = xp/q−1 y q −1 . . . . . .

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