I have been trying to wrap my head around related rates, which are super interesting but very difficult for me personally. Would anyone care to verify if my logic is correct here?
"A balloon rises into the air at 1.94 ft/sec. A girl letting go of the balloon runs 11.3 feet away. How fast is the angle of observation changing when the balloon is 18 feet from the ground?"
- Velocity, V, is 1.94 ft/sec so h = 1.94t
- t at 18 feet = 18 ft / 1.94 ft/sec = 9.28 seconds.
The equation we set up is between θ and the height, h:
- tan θ = opp/adj = h / 11.3
- differentiate both sides: (sec²(θ)) * d(theta)/d(t) = 1/11.3 * d(h)/d(t)
- Get dθ/dt alone on the left: dθ/dt = cos2θ * (1.94t/11.3)
- Cosine is adj/hyp, so cos2 becomes (11.3/sqr(11.32 + h2)) * 1.94t/11.3
- Solve to get the numerical answer for the the rate of change of the angle.
I've spent about 4 hours straight trying to work this out in my head, and even though I do understand implicit differentiation to a degree, I find this to be a whole different problem entirely! Thank you.