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 41E
To determine

To show: One family of the curves are orthogonal trajectories to the other family.

Expert Solution

Explanation of Solution

Given:

The family equations x2+y2=r2 and ax+by=0.

Derivative rules:

Chain rule: dydx=dydududx

Proof:

The equation of the curve is x2+y2=r2.

Differentiate implicitly with respect to x.

ddx(x2+y2)=ddx(r2)ddx(x2)+ddx(y2)=ddx(r2)2x+ddx(y2)=0

Apply the chain rule and simplify the terms,

2x+[ddy(y2)dydx]=02x+2ydydx=0dydx=xy

Thus, the slope of the tangent to the circle is dydx=xy.

Suppose that the circle and straight line are intersect at the point (x0,y0).

The slope of the tangent to x2+y2=r2 at (x0,y0) is m1=x0y0.

The given straight line is passing through point (0,0).

Clearly, the straight line passing through the points (x0,y0) and (0,0).

The slope of the tangent line computed as follows,

m2=y00x00=y0x0

Thus, the slope of the straight line is m2=y0x0.

Note: The two tangent lines are orthogonal if the product of their slopes is -1.

m1m2=(x0y0)(y0x0)=1

Hence, the family of the curves x2+y2=r2 and ax+by=0 are orthogonal trajectories is proved.

Graph:

The family of the circles and straight lines as shown below in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.5, Problem 41E

From Figure 1, it is observed that the family of the circle and family of the straight line are orthogonal trajectories.

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