# Math Help - Related rates into regards to applications of the derivative.

1. ## Related rates into regards to applications of the derivative.

Stuck on this one.
An airplane leaves a field at noon and flies east at 100 km/h and a second airplane leaves the same field at 1pm and flies south at 150 km/h. How fast are the airplanes separating at 2pm.

So if I do a right triangle.
My Vx right would be 200 and x would be 2 for hours//
My Vy(south) would be 150 and y 1 hour.
My Resultant of Vz is what I am looking for and z is 2.23606.
I am to find Vz and I come up with 245 km/h and the answer is 170 km/h.

Can someone see what I did wrong?? Thanks
Joanne

2. You need to pay attention to the sign +/- of the derivatives you're using. Velocities are vectors and vectors have directions. Did you draw a diagram and set up a coordinate system?

3. Hi, Our lessons did not include that and none of the questions I have answered, if - or + have to included.
Were just coming up with the number only. no direction for this question.

Hi, Our lessons did not include that and none of the questions I have answered, if - or + have to included.
Were just coming up with the number only. no direction for this question.
Oh, ok. Maybe you can avoid using vectors, and just look at it as a triangle. To find the position of the plane directed east, you need $\frac{dx}{dt}t$, where t is in hours. The distance of the second plane will be one hour behind, so it's position will be $\frac{dy}{dt}(t-1)$. The distance function could be written:

$s(t)=\sqrt{\left(\frac{dx}{dt}t\right)^2+\left[\frac{dy}{dt}(t-1)\right]^2}$

You're given the velocities:

$\frac{dx}{dt}=100$

$\frac{dy}{dt}=150$

So you plug those into the formula, and expand to obtain:

$s(t)=\sqrt{32,500t^2-45,000t+22,500}$

Use the chainrule to differentiate this function in order to find the rate at which the distance between the planes is increasing:

$\frac{ds}{dt}= \frac{2(32,500)t-45,000}{2\sqrt{32,500t^2-45,000t+22,500}}$

This derivative is equal to 170 at t=2.

5. Ok got it, I figured this out, and then did it another way how we should be doing it.
First by the Pyth to find z in kms, then doing a derivative of the x^2+ y^2=z^2.