#### 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).

#### Answer

The rate of change of the distance from the particle to the origin is dzdt=1+33 π2 cm/s.

#### Explanation

**Given:**

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

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

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

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

**Calculation:**

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

Then,

z=(x−0)2+(y−0)2 [QD=(x1−x2)2+(y1−y2)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=1310 cm

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

z2=x2+(2sin πx2)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)]

Put T=sin(πx2) in ddt[z2],

ddt[z2]=ddt[x2]+4ddt[T2] 2zdzdt=2xdxdt+8TdTdt [Q dydx=dydu⋅dudx]

Substitute T=sin(πx2) in the above equation,

2zdzdt=2xdxdt+8sin (πx2)ddx[sin πx2]dxdt2zdzdt=2xdxdt+8sin(πx2)×cos(πx2)×(π2)dxdtdzdt=1z[x+2πsin(πx2)×cos(πx2)]dxdt

Substitute x=13 and z=1310 and dxdt=10 in dzdt.

dzdt=11310[13+2πsin(π6)×cos(π6)]×10=3[13+2π(12)(32)]=33+33 π2=1+33 π2 cm/s

Thus, the rate of change of the distance from the particle to the origin is dzdt=1+33 π2 cm/s.