totallybord said:
Hi,
Help!
At noon a truck leaves Sydney dirving west at a speed of 40km/h. An hour later a 2nd truck leaves Sydney driving south at a speed of60km/h. At what rate is the distance btwn the 2 trucks increasing at 2pm that day?
Thanks for any help
What's the answer. Can you post it up. I want to see if I got it right. Then, I can post up my solution.
I will just do this:
Let x be the horizontal line and y be the vertical line and z between them.
So we have a right-angled triange.
dx/dt = 40 so x = 40t + 40 as this truck leaves an hour earlier so it already covered 40km
dy/dt = 60 so y = 60t (for this one, constant is zero because it starts from the origin)
For this question, we will be substituting t = 1 instead of 2 as the whole working-out has been adjusted to the car that is heading south, not the car that is going west.
z^2 = x^2 + y^2
= (40t+40)^2 + (60t)^2
= 1600t^2 + 3200t + 1600 + 3600t^2
= 5200t^2 + 3200 t + 1600
z = square root of (5200t^2 + 3200t + 1600)
dz/dt = 1/2 . 1/square root of (5200t^2 + 3200t + 1600) . (10400t + 3200) (substitute t = 1 here instead of t=2 as I adjusted the whole working-out for the car that went towards south, not the car that went west)
So dz/dt = 68km/h
I hope I am right...
