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 46E
To determine

To find: The value of a such that the family of the curves are orthogonal trajectories.

Expert Solution

Answer to Problem 46E

The curves are orthogonal trajectories when the value of a=33.

Explanation of Solution

Given:

The curves are y=(x+c)1 and y=a(x+k)13.

Derivative rules:

Chain rule: dydx=dydududx

Calculation:

Obtain the derivative of the curves.

Consider the curve y=(x+c)1.

Differentiate implicitly with respect to x,

ddx((x+c)1)=1(x+c)11ddx(x+c) (by chain rule)=1(x+c)11(1+0)=1(x+c)11=1(x+c)2

Thus, the derivative of the function y=(x+c)1 is 1(x+c)2.

That is, the slope of the tangent to the curve y=(x+c)1 is m1=1(x+c)2.

Consider the curve y=a(x+k)13.

Differentiate implicitly with respect to x,

ddx(a(x+k)13)=(a)(13)(x+k)131(1+0)=a3(x+k)23=a3(x+k)23

Thus the derivative of the function y=a(x+k)13 is a3(x+k)23.

That is, the slope of the tangent to the curve y=a(x+k)13 is m2=a3(x+k)23.

Since the curves are orthogonal trajectories, the product of its slope is -1.

m1m2=11(x+c)2(a3(x+k)23)=1a=3(x+c)2(x+k)23

Since y=(x+c)1 and y=a(x+k)13, y2=(x+c)2 and y2=a2(x+k)23,

a=3(1y2)(y2a2)a=3a2a3=3a=33

Therefore, the curves are orthogonal trajectories when the value of a=33.

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