Question

A stuntman falls from a hot-air balloon that is hovering at a constant altitude, 100 ft above a lake. A television camera on shore, 200 ft from a point directly below the balloon, follows his descent. At what rate is the angle of elevation, theta, of the camera changing 2 seconds after the stuntman falls? (Neglect the height of the camera.)

Answer 1

\[
\tan(\theta) = y/x
\] Implicitly differentiate with respect to \(t\). What equation do you get?

Answer 2

This?\[\sec ^{2}\theta \frac{ d \theta }{ dt } =\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }\]

Answer 3

You'd have to use the quotient rule actually.

Answer 4

\[
(y/x)' = \frac{y'x-yx'}{x^2}
\]

Answer 5

so then...\[\sec ^{2}\theta \frac{ d \theta }{ dt }=\frac{ \frac{ dy }{ dt }x-y \frac{ dx }{ dt } }{ x ^{2} }\]

Answer 6

Yeah, now since \(x\) is constant with respect to time, we know \(x'=0\).

Answer 7

Also, since \(x = \cos\theta\) we know \(\sec^2\theta = x^{-2}\).

Answer 8

This will simplify things quite a bit.

Answer 9

why is x=cosθ?

Answer 10

Ooops. I meant to say the \(z\cos\theta =x\) where \(z\) is the hypotenuse.

Answer 11

oh okay

Answer 12

So \(\sec^2\theta = z^2/x^2\)

Answer 13

\( z^2/x^2 =( y^2+x^2)/x^2\)

Answer 14

What equation are you ending up with for \(\theta'\)?

Answer 15

\[\frac{ d \theta }{ dt }=\frac{ \frac{ dy }{ dt }x }{ (y ^{2}+x ^{2}) }\]

Answer 16

Yeah. Now you are given y and x, all you need is y'

Answer 17

the gravitational constant -32.174ft/sec^2?

Answer 18

That's acceleration.

Answer 19

You want speed.

Answer 20

do you mean velocity such as taking the integral of 32.174ft/sec^2 and coming up with -32.174t ft/sec

Answer 21

\[
y' = \int_0^2y''dt
\]In this case, \(y''\) is the acceleration due to gravity.

Answer 22

so 64.384

Answer 23

Yeah. Now you can solve for \(\theta'\)

Answer 24

I mean you have everything you need now, since: \[
\theta' = \frac{y'x}{x^2+y^2}
\]

Answer 25

so...\[\frac{ d \theta }{ dt }=\frac{ 12869.6 }{ 50000 } = 0.257392\]

Answer 26

or no I forgot the negative sign

y'=-64.348

so... it should be -0.257392

correct?

Answer 27

It's in radians, I believe.

Answer 28

yes rads

Answer 29

but otherwise the answer is:
The rate of change in the angle of the camera at 2 seconds after the stuntman jumps is -0.257392rads

correct?

Answer 30

You could try differentiating: \[
z\sin\theta = y
\]to see if the answer is consistent.

Answer 31

actually that would be a bit complicated.

Answer 32

yes it seems like it would be :-)

Answer 33

Wait, actually there might be a problem with the answer you got.

Answer 34

We should probably use \(y(t=2)\) rather than \(y(t=0)\).

Answer 35

Why is that?

Answer 36

Because \(y=y(t)\) and we overlooked that fact.

Answer 37

It's a function of \(t\) and we set \(t=2\).

Answer 38

\[
\int_0^2y'dt = y(2)-y(0)\implies y(2) = \int_0^2y'dt + y(0)
\]

Answer 39

This time though the integrand (\(y'\) ) is not constant. \[
y'(t) = y''\cdot t= -32.174t
\]

Answer 40

Sorry I didn't notice that at first.

Answer 41

how is y'(t)=y'' * t =-32.174t?

Answer 42

it it because V=gt? where g=gravitational acceleration and t=time?

Answer 43

Okay so basically....\[
\int_0^u y''dt = y'(u) - y'(0)

\]By the fundamental theorem of calculus (\(u\) is just a dummy variable).
At the start, the velocity is 0, so \[
y(u) = \int_0^u y''dt = -32.174\int_0^udt =-32.174 \left[t \bigg|_0^u \right] = -32.174u
\]

Answer 44

Replacing \(t\) back into \(u\): \[
y'(t) = -32.174t
\]

Answer 45

Likewise: \[
y(2) -y(0)= \int_0^2 y'' dt \implies y(2) - 100 = \int^2_0 -32.174tdt
\]

Answer 46

The whole \(v=gt\) comes from calculus.

Answer 47

The derived formula are fine if you remember the preconditions for them, otherwise just use the definitions to get things done.

Answer 48

In math section, I avoid the derived formula and use mainly definitions / laws / principles.

Answer 49

got it

Answer 50

so it should be:\[\frac{ d \theta }{ dt }=\frac{ -32.174t*200 }{10000+40000 }\]

Answer 51

No, you never solved for \(y(2)\)

Answer 52

\[
\frac{d\theta}{dt}(2) = \frac{\frac{dy}{dt}(2)\cdot x(2)}{[y(2)]^2+[x(2)]^2}
\]

Answer 53

y(2)=-16.087t^2

Answer 54

You are evaluating when \(t=2\), there should be no \(t\)s left.

Answer 55

A definite integral always returns a number that is constant

Answer 56

so y(2)=-64.384

Answer 57

oh -64.348

Answer 58

You're saying they're underground? That makes no sense.

Answer 59

\[
y(2) -y(0)= \int_0^2 y'' dt \implies y(2) - 100 = \int^2_0 -32.174tdt
\]

Answer 60

\[
y(2) = 100 + -32.174\left[ \frac{t^2}{2}\bigg|_0^2\right]
\]

Answer 61

if y'=y''*t or -32.174t the integral of y' with limits of integration of [0,2] is -16.087{(2)^2}or -64.348

Answer 62

Yeah but you are forgetting that \(y(0) \neq 0 \)

Answer 63

it is = 100

Answer 64

Yeah.

Answer 65

so y(2) = 35.652

Answer 66

Now that makes sense.

Answer 67

so....\[\frac{ d \theta }{ dt }=\frac{ -64.348(200) }{ (35.653)^{2}+40000 }\]

Answer 68

so then θ' is -0.3118305217588708 or in answer form decreasing at a rate of 0.3118305217588708rads

Answer 69

Yeah that looks right.

Answer 70

okay, now I'm going to try and figure out were I went wrong the first time I tried to solve this. Thank you so much for your help :-)