   Chapter 3.9, Problem 24E

Chapter
Section
Textbook Problem

A particle moves along the curve y = 2 sin(πx/2). As the particle passes through the point ( 1 3 , 1 ) its x-coordinate increases at a rate of 10   cm / s . How fast is the distance from the particle to the origin changing at this instant?

To determine

To find: The rate of change of the distance from the particle to the origin when particle passes through the point (13,1).

Explanation

Given:

A particle moves along the curve y=2sin(πx2).

The rate of change of x co-ordinate is dxdt=10cm/s when (x,y)=(13,1).

Formula used:

(1) Chain rule: dydx=dydududx

(2) Distance between the two points (x1,y1) and (x2,y2):D=(x1x2)2+(y1y2)2.

Calculation:

Let z be the distance between the origin and the point (x,y) on the xy-plane.

Then,

z=(x0)2+(y0)2[QD=(x1x2)2+(y1y2)2]=x2+y2

Since x and y changes with the time t, x and y are the function of the time t.

Obtain dzdt when the particle passes through the point (13,1).

Distance of the particle from the origin when particle passes through the point (13,1).

z=(13)2+12=19+1=109=1310cm

Substitute y=2sin(πx2) in z and eliminates y from z.

z2=x2+(2sinπx2)2=x2+4sin2(πx2)

Differentiate z with respect to the time t,

ddt[z2]=ddt[x2+4sin2(πx2)]ddt[z2]=ddt[x2]+4ddt[sin2(πx2)

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