BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 25E
To determine

To find: The equation of the tangent line to the given equation at the point.

Expert Solution

Answer to Problem 25E

The equation of the tangent line to the equation x2+y2=(2x+2y2x)2 at (0,12) is y=x+12.

Explanation of Solution

Given:

The curve is x2+y2=(2x+2y2x)2.

The point is (0,12).

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x)  are both differentiable function, then

dydx=dydududx.

(2) Product rule: If y=f(u) and u=g(x)  are both differentiable function, then

ddx(f(x)+g(x))=ddx(f(x))+ddx(g(x)).

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

Where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1,y=y1.

Calculation:

Consider x2+y2=(2x+2y2x)2.

Differentiate the above equation implicitly with respect to x,

ddx(x2+y2)=ddx((2x+2y2x)2)ddx(x2)+ddx(y2)=ddx((2x+2y2x)2)2x+ddx(y2)=ddx((2x+2y2x)2)

Let u=2x2+2y2x and the chain rule (1),

2x+ddx(y2)=ddx(u2)2x+[ddy(y2)dydx]=ddu(u2)dudx2x+2ydydx=2ududx

Substitute u=2x2+2y2x,

2x+2ydydx=2uddx(2x2+2y2x)2x+2ydydx=2(2x2+2y2x)[ddx(2x2)+ddx(2y2)ddx(x)]2x+2ydydx=2(2x2+2y2x)[4x+2ddy(y2)dydx1]2x+2ydydx=2(2x2+2y2x)[4x+4ydydx1]

Substitute (0,12) for (x,y),

2(0)+2(12)dydx=2(0+2(12)20)[0+412dydx1]dydx=2(12)[2dydx1]dydx=2dydx1dydx=1

Thus, the slope of the tangent at (0,12) is m=1.

Substitute (0,12) for (x1,y1) and m=1 in equation (1),

y12=1(x0)y=x+12

Therefore, the equation of the tangent line to the equation x2+y2=(2x+2y2x)2 at (0,12) is y=x+12.

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