Question

Hey, the problem is below
http://postimg.org/image/5zy86as7n/

I will appreciate your help : )

Answer 1

Lucky you, studying at the University of Glasgow!!! My grandfather was born just a mile or two northeast of your classrooms, off Garscube Rd.
Suggest you draw a diagram to show clearly what your variables measure and how they are related to one another.

Answer 2

Note that we're dealing with a right triangle whose hypotenuse is s, vertical leg is x, and horizontal leg is 10 miles (fixed).

Answer 3

Once we have this relationship between s, x and 10 miles, we'll differentiate it with respect to time to find related rates. But that's a story for tomorrow.

Answer 4

I'm glad you're acquainted with the place : ) I've tried to get a formula for the change of s, namely s=sqrt(10^2+x^2), but I can't really figure out what to do after that. I know that dx/dt=30miles/h, but how do I connect this with the moment that 10 miles are left in the vertical direction? (if we go by your drawing)

Answer 5

It's 10 miles in the horiz. direction; this distance remains constant. The vertical distance between car and roundabout decreases as the car travels north, right?

I see you've already applied the Pyth. Thm. to obtain s=sqrt(100+x^2). Great.

Strictly speaking, you don't have to solve for s to carry out the next step, although you can if you wish.

So, supposing that you end up with s^2 = 10^2 + x^2,
take the derivative with respect to time, t, of this entire expression. Be certain to use the Chain Rule appropriately. Note that the derivative of 10^2 with respect to t will be zero. Perhaps this fact helps answer some of your question. Your goal is to find ds/dt.

Note that your vertical distance, x, is decreasing; therefore, dx/dt is -30 mph.

Give this a stab.

Altho' I need to hit the sack now, I'll most likely see in the morning anything further that you post. Good luck! We'll take a cup o' kindness yet, for Auld Lang Syne.

Answer 6

Haha, cheers, that was very helpful! : )

Answer 7

I got 30/sqrt(2). I wonder, since the horizontal distance does not contribute at all (it doesn't change), and at that moment (south-west) the car is seen at an angle of pi/4... So the rate of change can be velocity/sin(angle) or velocity/cos(angle)... I need to check for other values.

Answer 8

Just let me know if you have further questions. Cheers!