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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? |
Answer» Solution :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` |
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