# One family of the curves are orthogonal trajectories to the other family.

### Single Variable Calculus: Concepts...

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

### 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.

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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