# related rates question with pythagoras

I was doing this homework sheet and I could not solve this question.

2 aircrafts are flying at 530mph and at the same height along straight-line paths that are right angles to each other. Both aircrafts are flying towards the intersection part of the 2 paths. How fast is the distance between the 2 aircrafts decreasing when one aircraft is 6miles and the other aircraft is 9miles from the intersection part of the 2 paths?

To solve the question I drew a right angled triangle with lengths of 6 and 9. I am guessing from previous question I am trying to calculate the instantaneous rate of decrease on displacement (i.e. ds/dt) and it probably uses pythagoras if the two distances are perpendicular to each other.

Can someone please help me with using mathematical theories that are to complex for a non-maths student to understand. Thank you.

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You have asked $12$ question so far and have not accepted a single answer. Kindly consider accepting answers and read here (meta.stackexchange.com/questions/5234/…) for more details. – user17762 Oct 30 '12 at 20:43

It is I think not best to label the sides of the right-triangle $6$ and $9$. The sides should be labelled $x$ and $y$, for the sides of the triangle are changing.
We know that $\dfrac{dx}{dt}=\dfrac{dy}{dt}=-530$.
We want to find the rate of change of the distance. The distance $s$ between the aircraft is given by $$s=\sqrt{x^2+y^2}.$$ Now I will continue in not my favourite way. Differentiate $s$ with respect to $t$, using the Chain Rule. We get $$\frac{ds}{dt}=\frac{1}{2}\frac{2x\frac{dx}{dt}+2y\frac{dy}{dt}}{\sqrt{x^2+y^2}}.$$ Finally, freeze the situation at the instant when $x=6$ and $y=9$.
Another way: I would much prefer to note that $$s^2=x^2+y^2.$$ Now use implicit differentiation. We get $$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.\tag{1}$$ Much nicer! Finally, freeze the situation at the instant that $x=6$ and $y=9$. At that instant, $s=\sqrt{6^2+9^2}$. So at that instant, we know everything in Expression $(1)$ except $\dfrac{ds}{dt}$. We can therefore easily calculate $\dfrac{ds}{dt}$.