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 51E
To determine

To find: The points at which the slope of the curve is -1.

Expert Solution

Answer to Problem 51E

The slope of the curve is -1 at the points (1,1) and (1,1).

Explanation of Solution

Given:

The curve x2y2+xy=2.

Derivative rules:

Product rule: ddx(fg)=fddu(g)+gddx(f)

Chain rule: dydx=dydududx

Calculation:

Differentiate x2y2+xy=2 implicitly with respect to x,

ddx(x2y2+xy)=ddx(2)ddx(x2y2)+ddx(xy)=0

Apply the product rule,

[x2ddx(y2)+y2ddx(x2)]+[xddx(y)+yddx(x)]=0[x2ddx(y2)+y2(2x)]+[xdydx+y(1)]=0

Apply the chain rule and simplify the terms,

x2ddy(y2)dydx+2xy2+xdydx+y=0(2x2y+x)dydx+2xy2+y=0(2x2y+x)dydx=2xy2ydydx=2xy2y2x2y+x

Thus, the derivative of the equation is dydx=2xy2y2x2y+x.

That is, the slope of the tangent is dydx=2xy2y2x2y+x.

Obtain the required points such that the slope of the tangent is -1.

2xy2y2x2y+x=12xy2+y=2x2y+x2xy2+y2x2yx=02xy(yx)+yx=0

Simplify the expression,

(2xy+1)(yx)=0yx=0 or 2xy+1=0y=x or 2xy=1y=x or xy=12

Substitute xy=1 2 in the left hand side of the equation x2y2+xy=2,

x2y2+xy=(xy)(xy)+xy=(12)(12)+(12)=1412=14

That is, x2y2+xy2.

Therefore, the equation x2y2+xy=2 has no solution when xy=1 2.

Substitute y=x in the curve equation x2y2+xy=2,

x2(x)2+x(x)=2x4+x22=0

Let u=x2,

u2+u2=0(u+2)(u1)=0u=2 or u=1

Thus, the real solution are x2=1 and x=±1.

Substitute x=±1 in y=x, the points are (1,1) and (1,1)

Therefore, the slope of the curve is -1 at the point (1,1) and (1,1).

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