Question

Help Pleaseee!!!(:
Related Rates: A plane is flying horizontally at an altitude of 1 mi. and a speed of 500 mph passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi. away from the station.

Answer 1

Just from reading the problem and imagining the scenario in your head, you can figure out that the method of optimisation will be through pythagorean's theorem. Here is a drawing:As you can see that from the given situation, we are able to set up a distance relation using the pythagorean theorem. We express this as:\[\bf h^2+1^2=d^2 \implies h^2+1=d^2\]Differentiating both sides of the equation with respect to 't' - time we get:\[\bf 2h (h)'=2d (d)' \implies h(h)'=d(d)'\]Also note that the plane covers a horizontal distance of 500mi every hour, i.e. its horizontal speed with respect to time is 500mi/hr. Since speed is a ratio of distance over time, we can express this as:\[\bf \frac{ h }{ t }=500 \ mi/hr\]Differentiating both sides with respect to 't' - time, we get:\[\bf \frac{ (h)'t-h }{ t^2 }=0 \implies (h)'t-h=0 \implies \frac{ dh }{ dt }=\frac{ h }{ t }=500 \ mi/hr\]This is the same as what we started with. Note that if one is experienced with calculus, then we wouldn't have had to do this above calculations and may have immediately recognised that dh/dt = 500 instead of having to verify. But this may take some time and will come naturally as you do more problems =].

Now going back to the problem, we are asked to evaluate the change in the distance between the radar station and the plane when they are 2 mi away, i.e. we need to find
(d)' when d = 2. Going back to the equation we got earlier through differentiating both sides of the pythagorean relation, in order to find (d)', we must have 'd', and we know that d = 2. We also need to know (h)' which we also just deduced to be 500 mi/hr, but now we need to know 'h'. This is the horizontal distance of the triangle. We can find it by going back to the pythagorean triangle and plugging in d = 2 and solve for 'h':

\(\bf h^2+1=(2)^2 \implies h=\sqrt{3}\).

And now we have everything to solve for (d)'. We plug everything in and evaluate:\[\bf \sqrt{3}(500)=2(d)' \implies d'= \frac{ \sqrt{3}(500) }{ 2 }=250\sqrt{3} \approx 433.013 \ mi/hr\]Note that the units are "mi/hr".

Hence the distance between the radar station and the plane is increasing at a rate of \(\bf 433.013 \ mi/hr\) when \(\bf d=2\).

@pdd21