Calculus Related Rates Problems

  • Thread starter gina4930
  • Start date
  • #1
10
0
Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the figure). The point Q is on the floor h = 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 2.5 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q?
 

Answers and Replies

  • #2
10
0
I'm not too sure where to go with this but I think I got the start:

(x^2+12^2)^1/2+(y^2+12^2)^1/2=39

Can somewhere tell me if this is right and where to go from here?
 
  • #3
I like Serena
Homework Helper
6,577
176
Welcome to PF, gina4930! :smile:

You mentioned something about a figure?
I'm afraid it's too hard for me to deduce what you're doing without a figure. :redface:
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
961
I assume your picture shows two carts, on a level surface, connected by a rope passing over a pulley at some point above and between them. The height of the pulley is 12 ft. Let x be the distance from Q to one cart and y the distance from Q to the other cart. Then the hypotenuse of the right triangle formed by the first cart, P, and Q, is given by [itex]\sqrt{x^2+ 144}[/itex]. The hypotenuse of the right triangle formed by the second cart, P, and Q, is given by [itex]\sqrt{y^2+ 144}[/itex]. The two hypotenuses form the entire length of the rope: [itex]\sqrt{x^2+ 144}+ \sqrt{y^2+ 144}= 39[/itex]. That's the formula you have. Well, done!

That is the "static" formula. To get a "dynamic" formula, relating the rates of motion, differentiate both sides with respect to t, using the chain rule. Your final formula will involve both dx/dt and dy/dt. You are given one and asked to find the other.
 
  • #5
I like Serena
Homework Helper
6,577
176
Hey HoI! :wink:

That looks like a very plausible interpretation of the problem!
 
  • #6
10
0
When I differentiated the equation, I got:
1/2(x^2+144)^-1/2*(2x*dx/dt) + 1/2(y^2+144)^-1/2*(2y*dy/dt)=0
then I substituted in x=-5 , dx/dt= -2.5 , and y= (533)^1/2
When I solved for dy/dt I got two totally different answers. The first time I got -1.08 and the second time I got -.04. According to my online assignment, neither of these are correct. Can anyone point out my mistake? Thanks!
 
  • #7
I like Serena
Homework Helper
6,577
176
Looking good...

But how did you get y=(533)^1/2?
 
  • #8
10
0
I solved for y using the original equation. I thought using (533)^1/2 is more accurate then putting 23.1.
 
  • #9
I like Serena
Homework Helper
6,577
176
Ah well, after correcting a mistake of my own, I get y=√532, which is still different from yours.

And after that I get dy/dt=-1.084 m/s.

Since the problem asks how fast cart B is moving toward Q, the answer should be 1.084 m/s.
Can it be that it should be without a minus sign?
 
  • #10
10
0
I did not even think to put it without a minus sign. Thank you. Thank was the correct answer. I greatly appreciate it.
 
  • #11
I like Serena
Homework Helper
6,577
176
Good! :smile:
 

Related Threads on Calculus Related Rates Problems

  • Last Post
Replies
3
Views
1K
Replies
1
Views
5K
  • Last Post
Replies
11
Views
7K
  • Last Post
Replies
6
Views
32K
Replies
1
Views
263
Replies
1
Views
7K
Replies
4
Views
1K
Top