Question

So ya, this question is in relation to applications of derivatives, I'm just not sure how to solve. Any help would be awesome~

A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 130ft above the ground, a bicycle moving at a constant rate of 17ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 6 sec later?

Answer 1

dh/dt = 1
dx/dt=17

Answer 2

h is height
x represent position of bike, so dx reprent speed

Answer 3

now pythagorean triangle
x^2 + h^2 = d^2
\[d=\sqrt{h^2 + x^2}\]

Answer 4

Take partial from here

Answer 5

haven't done this for a while... so I probably made a mistake or two (or 3). but here's my attempt... if we differentiate the relation (pythagorean theorem above) we get 2(dh/dt)h+2(dx/dt)x=2d(dd/dt) and just sub in the values evaluated at t=6. h(6)=136, x(6)=17*6 then solve for dd/dt.

Answer 6

Yep, It looks right

Answer 7

yeah, it is better idea to take partial of d^2 =h^2 + x^2
gj

Answer 8

alright, i see how it works ands got an answer (sr about no reply earlier, net cut off -_-)... alright, thanks a mil u 2.