1. ## Related Rates Problems

Problem:
Two cars start moving from the same point. One travels north at 60 mph and the other travels west at 25 mph. At what rate is the distance between the cars increasing two hours later?

What I've Done:

So I drew a right triangle like this:

I know I want to find $\displaystyle \frac{dx}{dt}$ at t = 2, and I know I have $\displaystyle \frac{dN}{dt}$ which is 60mph and $\displaystyle \frac{dW}{dt}$ which is 25mph.

I wrote the equation $\displaystyle N^2 + W^2 = x^2$. Is this correct? Am I doing this right?

Do I just differentiate this equation with respect to x, and then plug in the values for $\displaystyle \frac{dN}{dt}$ and $\displaystyle \frac{dW}{dt}$ ?

I really have no idea what I'm doing. I think I need t in my equation somewhere...

Help?

2. Originally Posted by lysserloo
Problem:
Two cars start moving from the same point. One travels north at 60 mph and the other travels west at 25 mph. At what rate is the distance between the cars increasing two hours later?

What I've Done:

So I drew a right triangle like this:

I know I want to find $\displaystyle \frac{dx}{dt}$ at t = 2, and I know I have $\displaystyle \frac{dN}{dt}$ which is 60mph and $\displaystyle \frac{dW}{dt}$ which is 25mph.

I wrote the equation $\displaystyle N^2 + W^2 = x^2$. Is this correct? Am I doing this right?

Do I just differentiate this equation with respect to x, and then plug in the values for $\displaystyle \frac{dN}{dt}$ and $\displaystyle \frac{dW}{dt}$ ?

I really have no idea what I'm doing. I think I need t in my equation somewhere...

Help?
Differentiate you relationship with respect to t, and then sub in the proper values.

Hint:

Spoiler:
At $\displaystyle t=2,$ $\displaystyle x=\sqrt{120^2+50^2}$

3. So far so good.
Differentiate (implicitly) to find an expression for dx/dt.
Then plug in dN/dt=60, dW/dt=25, N=120 (which is the value of N when t=2), W=50 and x=sqrt(120^2+50^2).

+ not - under the square root, VonNemo19.

4. So I differentiated implicitly and got $\displaystyle \frac{dx}{dt} = (\frac{1}{2})*( \frac{(2N \frac{dN}{dt} + 2W \frac{dW}{dt})}{\sqrt{N^2 + W^2)}})$

Is this right?

I ended up getting the answer to be 65mph. Is that also right?

5. One travels north at 60mph, so after 2 hours N=120. You don't need a "formula" to work that out. Same idea for W.

Yes your derivative is correct. Now sub in the bits and pieces you know.

6. Did you get it?