# Related rates differentiation problem

1. Jul 27, 2012

### 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 $\frac{dx}{dt} = 35$ and $\frac{dy}{dt} = 25$ and $\frac{dZ}{dt} = ?$ I am just wondering on how to set up the equation.

Last edited: Jul 27, 2012
2. Jul 27, 2012

### eumyang

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

3. Jul 27, 2012

### frosty8688

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

4. Jul 27, 2012

### 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.

5. Jul 27, 2012

### frosty8688

Ok, thanks for the help.

6. Jul 27, 2012

### 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.

7. Jul 27, 2012

### frosty8688

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

8. Jul 27, 2012

### eumyang

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

9. Jul 27, 2012

### frosty8688

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

10. Jul 27, 2012

### 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 $(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}$ (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).

11. Jul 27, 2012

### frosty8688

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

12. Jul 27, 2012

### eumyang

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

13. Jul 27, 2012

### frosty8688

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

14. Jul 27, 2012

### 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.

15. Jul 27, 2012

### frosty8688

What I got is $\frac{4300}{10\sqrt{101}}$ km/hr

16. Jul 27, 2012

### frosty8688

17. Jul 27, 2012

### eumyang

Last edited: Jul 27, 2012
18. Jul 27, 2012

### frosty8688

35 * 4 = 140 for x, 25 * 4 = 100 for y; $\frac{-2(150-140)*35+2*100*25}{\sqrt{(150-140)^{2}+ 100^{2}}}$ = $\frac{-2*10*35 + 5000}{\sqrt{10^{2} + 100^{2}}}$ = $\frac{-700 + 5000}{\sqrt{10100}}$ = $\frac{4300}{10\sqrt{101}}$

19. Jul 27, 2012

### eumyang

Looks like you removed the 2 in the denominator, but never factored out or canceled the 2's in the numerator. Let's just forget about canceling out the 2. You should have written:
$$\frac{-2(150-140) \cdot 35+2 \cdot 100 \cdot 25}{2\sqrt{(150-140)^{2}+ 100^{2}}}$$
(Also, don't use "*" for multiplication in LaTeX. Use "\cdot" instead.)

20. Jul 27, 2012

### frosty8688

I see, my mistake. I must have missed the 2 in the denominator.