Change in camera angle viewing rocket

[Gascoigne]

A rocket is rising according to the equation s=50t^2. A person at ground level is filming the rocket 2000 feet away. Rate of change in angle of elevation?

I thought I could do this problem like this:

But the answer doesn't doesn't match any of the ones I have to pick from. I know they're related rates, but I don't know how to solve it that way.

 

[galactus]

\(\displaystyle tan{\theta}=\frac{s}{2000}\)

But \(\displaystyle s=50t^{2}, \;\ \frac{ds}{dt}=100t\)

When t=10, s=5000. By Pythagoras, we see the hypoteneuse is \(\displaystyle 1000\sqrt{29}\)

\(\displaystyle sec^{2}{\theta}\frac{d{\theta}}{dt}=\frac{1}{2000}\frac{ds}{dt}\)

When x=2000, then \(\displaystyle sec{\theta}=\frac{\sqrt{29}}{2}\)

\(\displaystyle (\frac{\sqrt{29}}{2})^{2}\frac{d{\theta}}{dt}=\frac{1}{2}\)

\(\displaystyle \frac{d{\theta}}{dt}=\frac{2}{29}\approx .069 \;\ rad/sec\)

Maybe I made a mistake. Check it out.

Most of the time they give you a rate at which the rocket is rising. In this event they give you a position at some time t.

So, to find the velocity, we differentiate the position function, s

 

[Deleted member 4993]

A rocket is rising according to the equation s=50t^2. A person at ground level is filming the rocket 2000 feet away. Rate of change in angle of elevation?

I thought I could do this problem like this:

But the answer doesn't doesn't match any of the ones I have to pick from. I know they're related rates, but I don't know how to solve it that way.

\(\displaystyle \theta \, = \, \tan^{-1}(\frac{t^2}{40})\)

\(\displaystyle \frac{d\theta}{dt} \, = \, \tfrac{1}{1 + (\frac{t^2}{40})^2}\cdot\frac{2t}{40}\)

Now continue....

 

[Gascoigne]

A rocket is rising according to the equation s=50t^2. A person at ground level is filming the rocket 2000 feet away. Rate of change in angle of elevation?

I thought I could do this problem like this:

But the answer doesn't doesn't match any of the ones I have to pick from. I know they're related rates, but I don't know how to solve it that way.

\(\displaystyle \theta \, = \, \tan^{-1}(\frac{t^2}{40})\)

\(\displaystyle \frac{d\theta}{dt} \, = \, \tfrac{1}{1 + (\frac{t^2}{40})^2}\cdot\frac{2t}{40}\)

Now continue....

I actually included that in my solution, but forgot to put it in the explanation. Still wrong.

 

[Deleted member 4993]

show calculations...

 

[Gascoigne]

show calculations...

1/(1+100/40)^2 = 1/(7/2)^2 = 1/(49/4) = 49/4 * 1/2 = 2/49
galactus' answer is the one listed. I'm just losing track of the thought process halfway through...

\(\displaystyle tan{\theta}=\frac{s}{2000}\)

But \(\displaystyle s=50t^{2}, \;\ \frac{ds}{dt}=100t\)

When t=10, s=5000. By Pythagoras, we see the hypoteneuse is \(\displaystyle 1000\sqrt{29}\)

\(\displaystyle sec^{2}{\theta}\frac{d{\theta}}{dt}=\frac{1}{2000}\frac{ds}{dt}\)


I get everything up to this. Where does the secant value come from?
When x=2000, then \(\displaystyle sec{\theta}=\frac{\sqrt{29}}{2}\)

\(\displaystyle (\frac{\sqrt{29}}{2})^{2}\frac{d{\theta}}{dt}=\frac{1}{2}\)

\(\displaystyle \frac{d{\theta}}{dt}=\frac{2}{29}\approx .069 \;\ rad/sec\)

 

[Deleted member 4993]

show calculations...

1/(1+100/40)^2 <<< How is that???

The above should be:

\(\displaystyle \frac{1}{1+(\frac{100}{40})^2}\cdot\frac{2\cdot 10}{40}\)

Now finish it.....


= 1/(7/2)^2 = 1/(49/4) = 49/4 * 1/2 = 2/49
galactus' answer is the one listed. I'm just losing track of the thought process halfway through...

\(\displaystyle tan{\theta}=\frac{s}{2000}\)

But \(\displaystyle s=50t^{2}, \;\ \frac{ds}{dt}=100t\)

When t=10, s=5000. By Pythagoras, we see the hypoteneuse is \(\displaystyle 1000\sqrt{29}\)

\(\displaystyle sec^{2}{\theta}\frac{d{\theta}}{dt}=\frac{1}{2000}\frac{ds}{dt}\)


I get everything up to this. Where does the secant value come from?
When x=2000, then \(\displaystyle sec{\theta}=\frac{\sqrt{29}}{2}\)

\(\displaystyle (\frac{\sqrt{29}}{2})^{2}\frac{d{\theta}}{dt}=\frac{1}{2}\)

\(\displaystyle \frac{d{\theta}}{dt}=\frac{2}{29}\approx .069 \;\ rad/sec\)

 

[galactus]

Sec is hypoteneuse over adjacent. \(\displaystyle \frac{1000\sqrt{29}}{2000}=\frac{\sqrt{29}}{2}\)

The hypoteneuse came from Pythagoras knowing that x=2000 and s=5000

 

[Gascoigne]

[quote="Subhotosh Khan":u6molv94]show calculations...

1/(1+100/40)^2 <<< How is that???

The above should be:

\(\displaystyle \frac{1}{1+(\frac{100}{40})^2}\cdot\frac{2\cdot 10}{40}\)

Now finish it.....


= 1/(7/2)^2 = 1/(49/4) = 49/4 * 1/2 = 2/49
galactus' answer is the one listed. I'm just losing track of the thought process halfway through...

\(\displaystyle tan{\theta}=\frac{s}{2000}\)

But \(\displaystyle s=50t^{2}, \;\ \frac{ds}{dt}=100t\)

When t=10, s=5000. By Pythagoras, we see the hypoteneuse is \(\displaystyle 1000\sqrt{29}\)

\(\displaystyle sec^{2}{\theta}\frac{d{\theta}}{dt}=\frac{1}{2000}\frac{ds}{dt}\)


I get everything up to this. Where does the secant value come from?
When x=2000, then \(\displaystyle sec{\theta}=\frac{\sqrt{29}}{2}\)

\(\displaystyle (\frac{\sqrt{29}}{2})^{2}\frac{d{\theta}}{dt}=\frac{1}{2}\)

\(\displaystyle \frac{d{\theta}}{dt}=\frac{2}{29}\approx .069 \;\ rad/sec\)

[/quote:u6molv94]
...Oh. Getting that right would have helped.

Sec is hypoteneuse over adjacent. \(\displaystyle \frac{1000\sqrt{29}}{2000}=\frac{\sqrt{29}}{2}\)

The hypoteneuse came from Pythagoras knowing that x=2000 and s=5000

Thanks

The problem is that I understand the method after it's been explained, but I can't model it myself. For instance:
Snow is piled onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 15 feet high?

d=3h
V=1/3*pi*r^(2)h
=1/3*pi*9/4*h^3
=3/4*pi*h^3
?

V=1/3*pi*r^(2)h
V=1/3*pi*r^(2)*2/3*r
= 2/9*pi*r^(2)r
=2/9*pi*r^(3)
?

I don't know see how to relate the related rates, so I'm always stuck with at least one extra variable.

 

[galactus]

Since the diameter is 3 times the height, we have d=3h.

Use \(\displaystyle 2r=3h\) to eliminate r in the cone formula.

Now, do your thing. You will solve for dh/dt. Everything else is known.

When doing related rates, label everything you're given and label what you need.

In this case we are given dV/dt, r=3h/2, h. We want dh/dt.

Knowing we need dh/dt tells us that we need it in terms of h. So, eliminate r. That is where the condition 'diameter is 3 times the height comes in.

If they would have asked for dr/dt instead, we would have eliminated h. See?.

 

[Gascoigne]

Since the diameter is 3 times the height, we have d=3h.

Use \(\displaystyle 2r=3h\) to eliminate r in the cone formula.

Now, do your thing. You will solve for dh/dt. Everything else is known.

When doing related rates, label everything you're given and label what you need.

In this case we are given dV/dt, r=3h/2, h. We want dh/dt.

Knowing we need dh/dt tells us that we need it in terms of h. So, eliminate r. That is where the condition 'diameter is 3 times the height comes in.

If they would have asked for dr/dt instead, we would have eliminated h. See?.

No, I don't.

v=(1/2)(pi)(r^2)(h)
V = (9/8)(pi)(h^2)

dv/dt = (9/4)(pi)(h)(dh/dt)
...?

You said we're given dv/dt - where?

 

[galactus]

Snow is piled onto a conical pile at a rate of 10 cubic feet per minute

.

Right here.

dV/dt=10, we want dh/dt when h=15

\(\displaystyle 2r=3h\)

\(\displaystyle r=\frac{3h}{2}\)

\(\displaystyle V=\frac{\pi}{3}(\frac{3h}{2})^{2}h\)

Now, finish up by differentiating, plugging in the knowns and solving for dh/dt.

 

[Gascoigne]

Snow is piled onto a conical pile at a rate of 10 cubic feet per minute

.

Right here.

dV/dt=10, we want dh/dt when h=15

\(\displaystyle 2r=3h\)

\(\displaystyle r=\frac{3h}{2}\)

\(\displaystyle V=\frac{\pi}{3}(\frac{3h}{2})^{2}h\)

Now, finish up by differentiating, plugging in the knowns and solving for dh/dt.

The with-respect-to-time part throws me off.

edit: Answer = 8/(405(pi))
Would I apply the same technique to this?
A snowball melts at a rate of 10 cubic cm per minute. At what rate is the radius changing when the snowball is 1 cm in radius?
Have: dv/dt
Looking for: dr/dt

v = (4/3)(pi)(r^3)

dv/dt = -10

-10 = 12(pi)(r^2)(dr/dt)
dr/dt = -10/(12(pi)(r^2) = -.265

That looks right to me, but I don't have that available as an answer. Sorry, I'm just having such a hard time with this stuff.

edit 2: stupid mistakes. Got the answer.