[Math] Related Rates and Angle of elevation

calculusderivatives

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?"

Given:

  • Velocity, V, is 1.94 ft/sec so h = 1.94t
  • t at 18 feet = 18 ft / 1.94 ft/sec = 9.28 seconds.

picture of problem

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.

Best Answer

That's incorrect because the rate of change of height is already given to be 1.94, you don't need to solve for t. I mean you could, but as I'll show, its unnecessary.

Let's start by writing what we know. Let h=height of balloon, x=ground distance from balloon,

and let $\theta$=angle

$${dh \over dt}=1.94$$ $$x=11.3$$

Time to use some trigonometry, note that you want the angle, specifically the angle of elevation. Also note that you have both the h height and x distance. So you have a triangle, with the length adjacent to the angle given, and the length opposite to the angle given at a point in time. SOH CAH TOA, and you'll find you need the tangent. You already seem to know that part. $$tan{(\theta)}=h/x$$ $$\Rightarrow \theta=arctan(h/x)$$ Now, h is actually a function of t so we'll write h(t) instead. Now, lets take the derivative of $\theta$ with respect to h(t). We'll use the chain rule for h(t). Also note ${d \over du}arctan(u/a)={a \over u^2+a^2}$ $${d{\theta} \over d{h(t)}}={dh \over dt}{x \over (h(t))^2+x^2}$$ Cancel the x's and substitute resulting in $${1.94*11.3 \over 18^2+11.3^2}$$ So the rate of change of the angle of elevation when the balloon is 18 feet high is approximately equal to 0.0485 radians per second.

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