#### To determine

**To find:** The rate of change of the distance from the plane to the radar station.

#### Answer

The rate of change of the distance between the plane and radar station is dydt≈296 km/h.

#### Explanation

**Formula used:**

Chain rule: dydx=dydu⋅dudx.

**Calculation:**

Let us assume that at any time *t*, *A* be the position of top of radar station on the ground and *P* be the position of the plane 1 minute later after crossing the radar and *B* be the position of the radar on the ground.

Let *x* be the distance between the top of the radar and plane and *y* be the distance between the ground point of the radar and the plane as shown in the Figure 1 given below.

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

Since the plane climbs at an angle of 30° from the horizontal, the total angle between the plane and radar is 120°.

Obtain dydt when t=1 min later.

From the cosine rule in the ΔAPB.

y2=12+x2−2(1)(xcos(120))=1+x2−2x(−12)=1+x+x2

Differentiate y2=1+x+x2 with respect to the time *t*.

ddt(y2)=ddt(x2+x+1)2ydydt=(2x+1)dxdt [∵dydx=dydu⋅dudx]dydt=(2x+12y)dxdt

After 1 minute later distance travelled by the plane.

x=30060=5 km

When x=5, the value of *y* is,

y=1+52+5=1+25+5=31

Substitute x=5, y=31 and dxdt=300 in dydt,

dydt=(2(5)+1231)(300)=11231(300)=11(150)31=165031≈296 km/h

Therefore, the rate of change of the distance between the plane and radar station is dydt≈296 km/h.