Question

Inverse trig differentiation
A television camera at ground level is filming the lift-off of a space shuttle at a point 750 meters from the launch pad. Let theta be the angle of elevation of the shuttle and let s be the distance between the camera and the shuttle (see figure attached). Write as theta a function of s for the period of time when the shuttle is moving vertically. Differentiate the result to find dtheta/dt in terms of ds/dt and s.

Answer 1

\[\sec{\theta} =\frac{s}{750}\]\[\theta =\sec^{-1}{\frac{s}{750}}\]\[\frac{d\theta}{ds}=\frac{750}{s\sqrt{s^2-750^2}}\]

Answer 2

http://www.themathpage.com/acalc/inverse-trig.htm#arcsec

Answer 3

Thanks this chapter has been tripping me up... that's it?

Answer 4

No, you need d theta/dt.

Answer 5

Okay, so how would I differentiate for /dt if dtheta is currently in terms of ds

Answer 6

Well \[s=\sqrt{h^2+750^2}\]Let h=h(t). h(t) will be some function which is at minimum more straightforward (with h(0)=0). \[s(t) =\sqrt{h(t)^2+750^2}\]\[\frac{ds}{dt} =\frac{d\sqrt{h(t)^2+750^2}}{dt}=\frac{h(t)h'(t)}{\sqrt{h(t)^2+750^2}}\]

Answer 7

what rule did you apply to get the derivative? I'm kind of lost

Answer 8

\[f(t) = (g(t)+c)^{1/2}\]\[f'(t) = \frac{1}{2}(g(t)+c)^{-1/2}g'(t)\]

Answer 9

Oh, okay

Answer 10

\[\frac{d\theta}{dt}=\frac{d\theta}{ds}*\frac{ds}{dt}\]With some cool algebraic repacements for s I think.

Answer 11

I know I sound stupid but that's so cool! It's like dimensional analysis for derivatives lol

Answer 12

Let me try and do this and I'll put up my answer and you confirm

Answer 13

Wait, I think I have it

Answer 14

Or maybe you don't need to do algebraic replacements. It might be enough to give "dtheta/dt in terms of ds/dt and s."

Answer 15

Looks like you have d theta /ds upside down.

Answer 16

Plus, I don't actually think you need to add h(t) into the mix after all.

Answer 17

Okay, thank you