A police jeep, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the jeep is `0.6km` north of the intersection and the car is `0.8km` to the east, the police determine with radar that the distance between them and the car is increasing at `20kmh^-1`. If the jeep is moving at `60kmh^-1` at the instant of measurement, what is the speed of the car?

Explanation

We draw a diagram of the car and jeep in the coordinate plane, using the positive x-axis as the eastbound highway and the positive y-axis as the northbound highway(figure). Let x be the position of car at time t.

y=position of jeep at time t,
s=distance between can and jeep at time t.
We assume x,y and s to be differentiable functions of t.
`x=0.8km`, `y=0.6km`, `(dy)/(dt)=-60kmh^-1`
`(ds)/(dt)=20kmh^-1` (`dy//dt` is negative because y is decreasing.)
The variables are related as:
`s^2=x^2+y^2` ...(i) (Pythagorean theorem)
Differentiate Eq. (i) with respect to t, we get
`(ds^2)/(dt)=(dx^2)/(dt)+(dy^2)/(dt)=(ds^2)/(ds)*(ds)/(dt)=2s(ds)/(dt)`
Similarly, `(dx^2)/(dt)=(dx^2)/(dx)*(dx)/(dt)=2x(dx)/(dt)`
and similarly `(dy^2)/(dt)=(dy^2)/(dy)*(dy)/(dt)=2y(dy)/(dt)`
`2s(ds)/(dt)=2x(dx)/(dt)+2y(dy)/(dt)`
Chain rule, `(ds)/(dt)=1/s(x(dx)/(dt)+y(dy)/(dt))`
`=(1)/(sqrt(x^2+y^2))(x(dx)/(dt)+y(dy)/(dt))`
Evaluate, with `x=0.8km`, `y=0.6km`, `dy//dt=-60kmh^-1`,
`ds//dt=20kmh^-1`, and solve for `dx//dt`.
`20=(1)/(sqrt(underbrace((0.8)^2+(0.6)^2)))(0.8(dx)/(dt)+(0.6)(-60))`
`implies 20=0.8(dx)/(dt)-36implies(dx)/(dt)=(20+36)/(0.8)=70`