Related rates differentiation problem

frosty8688

1. At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4:00 pm?



2. None



3. I have the distance as 150 km. I have the variables [itex]\frac{dx}{dt} = 35[/itex] and [itex]\frac{dy}{dt} = 25[/itex] and [itex]\frac{dZ}{dt} = ?[/itex] I am just wondering on how to set up the equation.

eumyang

What's the relationship between x, y, and Z? Note that Z is the hypotenuse of a right triangle.

frosty8688

x is ship A's position, y is ship B's position and Z is the rate of change in distance.

Vorde

Realize that the distance equation is the Pythagorean equation, so try setting up a triangle with what you know, and see if you can figure out how to get the hypotenuse of the triangle from what you are given.

frosty8688

Ok, thanks for the help.

eumyang

Oops, I misread the part that stated that A started west of B and is traveling east. Suppose we call the point where ship B started point O. |AO| = 150. As time goes on, A gets closer to O. What expression would give us the distance |AO| in terms of x?

Also, Z is the distance between A and B. It is not the rate of change in distance.

frosty8688

The equation would be: [itex]\frac{dz}{dt}[/itex] = [itex]\frac{(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}}{2z}[/itex]

eumyang

No. First, the distance |AO| would be 150 - x, not 150 + x. 2nd, show us the equation BEFORE taking the derivative.

frosty8688

The equation before taking the derivative is (150 - x)[itex]^{2}[/itex] + y[itex]^{2}[/itex] = z[itex]^{2}[/itex]

eumyang

I think I see what you did earlier. On the side with the x and y, take the derivative separately, and put the dx/dt and dy/dt next to each term. Earlier, you wrote [itex](2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}[/itex] (ignoring the 150+x error), which makes no sense. The numerator should be written as
(derivative of (150 - x)²)(dx/dt) + (derivative of y²)(dy/dt).

frosty8688

It would be [itex]\frac{dz}{dt}[/itex] = [itex]\frac{(150 - x)\frac{dx}{dt} + y \frac{dy}{dt}}{z}[/itex]

eumyang

Nope. You didn't take the derivative of (150 - x)² correctly. Nor the derivative of y².

frosty8688

The derivative would be [itex]\frac{2(150-x)\frac{dx}{dt} + 2y\frac{dy}{dt}}{2z}[/itex]

eumyang

Close. You need a negative in front of the 2(150 - x)(dx/dt). (You need to use the chain rule.) Also, to simplify things, factor out the 2 in the numerator and cancel with the 2 in the denominator.

Find the distance ship A traveled (x) in 4 hours. Find the distance ship B traveled (y) in 4 hours. Find Z. Then plug everything in the dZ/dt equation.

frosty8688

What I got is [itex]\frac{4300}{10\sqrt{101}}[/itex] km/hr

frosty8688

Which is about 42.79 km/hr

eumyang

That's not what I got. Can you please show all of your work here, instead of just writing the final answer?