## Related rates differentiation problem

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 $\frac{dx}{dt} = 35$ and $\frac{dy}{dt} = 25$ and $\frac{dZ}{dt} = ?$ I am just wondering on how to set up the equation.
 Recognitions: Homework Help What's the relationship between x, y, and Z? Note that Z is the hypotenuse of a right triangle.
 x is ship A's position, y is ship B's position and Z is the rate of change in distance.

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## Related rates differentiation problem

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.
 Ok, thanks for the help.

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 Quote by frosty8688 x is ship A's position, y is ship B's position and Z is the rate of change in distance.
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.
 The equation would be: $\frac{dz}{dt}$ = $\frac{(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}}{2z}$

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 Quote by frosty8688 The equation would be: $\frac{dz}{dt}$ = $\frac{(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}}{2z}$
No. First, the distance |AO| would be 150 - x, not 150 + x. 2nd, show us the equation BEFORE taking the derivative.
 The equation before taking the derivative is (150 - x)$^{2}$ + y$^{2}$ = z$^{2}$

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 Quote by frosty8688 The equation before taking the derivative is (150 - x)$^{2}$ + y$^{2}$ = z$^{2}$
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 $(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}$ (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).
 It would be $\frac{dz}{dt}$ = $\frac{(150 - x)\frac{dx}{dt} + y \frac{dy}{dt}}{z}$

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 Quote by frosty8688 It would be $\frac{dz}{dt}$ = $\frac{(150 - x)\frac{dx}{dt} + y \frac{dy}{dt}}{z}$
Nope. You didn't take the derivative of (150 - x)2 correctly. Nor the derivative of y2.
 The derivative would be $\frac{2(150-x)\frac{dx}{dt} + 2y\frac{dy}{dt}}{2z}$

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 Quote by frosty8688 The derivative would be $\frac{2(150-x)\frac{dx}{dt} + 2y\frac{dy}{dt}}{2z}$
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.
 What I got is $\frac{4300}{10\sqrt{101}}$ km/hr
 Quote by frosty8688 What I got is $\frac{4300}{10\sqrt{101}}$ km/hr