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 39E
To determine

To find: The points on the lemniscate where the tangent is horizontal.

Expert Solution

Answer to Problem 39E

The points on the lemniscate where the tangent is horizontal are (±534,±54).

Explanation of Solution

Given:

The equation is 2(x2+y2)2=25(x2y2) (1)

Derivative rule: Chain rule

If y=f(u) and u=g(x)  are both differentiable function, then dydx=dydududx.

Calculation:

Obtain the point at which the tangent line is horizontal.

Consider the equation 2(x2+y2)2=25(x2y2).

Differentiate implicitly with respect to x,

ddx(2(x2+y2)2)=ddx(25(x2y2))2ddx((x2+y2)2)=25ddx((x2y2))

Let u=x2+y2 and v=x2y2.

2ddx(u2)=25ddx(v)

Apply the chain rule (1) and simplify the terms,

2(ddu(u2)dudx)=25dvdx2(2ududx)=25dvdx4ududx=25dvdx

Substitute u=x2+y2 and v=x2y2,

4(x2+y2)ddx(x2+y2)=25ddx(x2y2)4(x2+y2)[ddx(x2)+ddx(y2)]=25(ddx(x2)ddx(y2))4(x2+y2)[2x+ddy(y2)dydx]=25(2xddy(y2)dydx)4(x2+y2)[2x+2ydydx]=25(2x2ydydx)

Combine the terms dydx to one side of the equation,

8x(x2+y2)+8y(x2+y2)dydx=50x50ydydx(8y(x2+y2)+50y)dydx=50x8x(x2+y2)dydx=50x8x(x2+y2)8y(x2+y2)+50y

Thus, the derivative of the equation is dydx=50x8x(x2+y2)8y(x2+y2)+50y.

Note that, the tangent is horizontal if dydx=0.

50x8x(x2+y2)8y(x2+y2)+50y=050x8x(x2+y2)=08x(x2+y2)=50xx2+y2=254

Substitute x2+y2=254 in equation (1),

2(254)2=25(x2+y22y2)2(254)2=25(2542y2)2528=252450y250y2=252(1418)

Simplify the terms and obtain the value of y,

50y2=252(1418)y2=252(18)y2=2516y=±54

Substitute y2=2516 in x2+y2=254,

x2+y2=254x2=2542516x2=7516x=±534

Therefore, the tangent line is horizontal at the points (±534,±54).

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