Question

A balloon rises at the rate of 8 ft/s from a point on the ground 60 ft from an observer. Find the rate of change of the angle of elevation when the balloon is 25 ft above the ground.

Answer 1

You're given opposite and adjacent. which means you're using tangent.

Answer 2

\[\tan (\theta) = \frac{ h }{ 60 }\]

Answer 3

i thought it was this...

Answer 4

solve for c using pythagorean theorem.

Answer 5

uh huh. that's 65

Answer 6

I don't think you need more than the tangent function to solve this one.

Answer 7

isn't it related rates tho? so u need to find d theta/dt?

Answer 8

We are told that
\[\frac{ dh }{ dt } = 8\]
\[\sec^2(x) \frac{ dh }{ dt } = \frac{ 8 }{ 60 } =\frac{ 2 }{ 15 }\]
divide by sec (or multiply by cos)
\[\frac{ d \theta }{ dt } = \frac{ 2 }{ 15 } \cos^2(x)\]

Answer 9

Yes, but the pythagorean theorem isn't useful here.

Answer 10

Wait, maybe I'm doing it differently, but my stuff doesn't look like abbot's, but kind of similar. I'll just spit it out give me a sec.

Answer 11

:)

Answer 12

By the Pythagoreas
\[c = \sqrt{25^2 + 60^2} = 65\]
and
\[\cos(x) = \frac{ 60 }{ 65 } = \frac{ 12 }{ 13 }\]

Answer 13

\[\tan \theta = \frac{ x }{ 60 }\]
so sec^2 θ (dθ/dt)= 1/60
so find θ using tan^-1θ = 25/60; θ = 22.62
so sec^2(22.62) dθ/dt = 1/60
so....
dθ/dt = 1/15 ?

Answer 14

x, y, and dy/dt are given, theta is simply tan^-1(y/x)

I just did the derivative of:

\[\tan \theta =y*x^{-1}\]

\[\sec^2 \theta \frac{ d \theta }{ dt }=\frac{ dy }{ dt }x^{-1}-x^{-2}\frac{ dx }{ dt }\]

since x isn't changing, dx/dt=0 and the final term falls off and we can plug in for everything else to solve for theta.

Answer 15

\[\frac{ d \theta }{ dt } = \frac{2}{15} \frac{12}{13}2 \approx \]

Answer 16

@Kainui for the θ u find it right? using tan θ = 25/60?

Answer 17

Yep, just take arctan to both sides and you get theta that way. It's basically given to you from the problem.

Answer 18

calculator in radian mode right? because it's in degree mode right now and i'm getting the wrong answer.

Answer 19

Yeah, the derivative of the tangent function in degrees is not secant squared.

Answer 20

kk thanks :D