Question

Related Rates

Finding the rate of change in height of a rising balloon.

Triangle xhz with z being the hypotenuse.
x = 500ft h=unkown z=irrelevant.
Angle theta between x and z =Pi/4
Change in theta = 0.14rad/min

So I need an explanation as to why my Professor got an answer that is double than mine. Here is my working.

dH/dt =? dx/dt = 0 dz/dt=?

y=tan(theta)=Opp/Adj = H/x
dy/dx = {[sin(theta)/cos(theta)]*d(theta)/dt} = dH/dt * 500^(-1)

Therefore
Note [sin(Pi/4)/cos(Pi/4)] = 1

dH/dt = 1* 0.14 * 500
dH/dt = 70ft/min ascending.

However my professors answer is 140ft/min

Answer 1

He works out [sin(theta)/cos(theta)] being 1 as me, but than takes pythag and substitutes that into the formula.

This becomes Sqrt(2) * 0.14 *500 and he gets 140ft / min. I dont understand why he would do that.

Answer 2

Your expression for dy/dx is incorrect because you need to take the derivative of tan and multiply it by the time derivative of the angle. I gather from the problem that x is fixed at 500 and so here's how I'd do it:
\[h=500tanx \rightarrow \frac{ dh }{ dt }=\sec^2 \theta \frac{ d \theta }{ dt } \]
But the square of cos(pi/4) is 1/2 and so we get:
\[ \frac{ dh }{ dt }=\frac{ 500 }{ \cos^2(\pi/4) } \frac{ d \theta }{ dt }=500(2)(0.14)\]

Answer 3

dy/dx of tan = sinx/cosx or sec^2x as you said. Both evaluate to 1. Right so you substituted sec^2x as 1/cos^2 x. ok.

That answer makes sense, however, my method should also work as I used the correct derivative of tanx?

Answer 4

oh bummer, I have them the wrong way around

Answer 5

hmm makes no difference tho, still comes out to 1.....

Answer 6

These identities are equal to each other so I don't understand why my method does not work.

Answer 7

By the way I forgot the 500 in front of the right side of the equation for dy/dx. It's late for me and I forgot to point out that you wrote dy/dx when I assume you meant dy/dt. What you wrote is then equivalent to:
\[\frac{ dy }{ dt }=\frac{ \sin \theta }{ \cos \theta }\frac{ d \theta }{ dt }= \tan \theta \frac{ d \theta }{ dt }\]
The tan should be sec^2.

Answer 8

What I meant is dy/dt, but does that make a difference?

Answer 9

Since y=h/x=h/500 we should have:
\[\frac{ dy}{ dt }=\frac{ 1 }{ 500 }\frac{ dh}{ dt }=\sec^2 \theta\frac{ d \theta }{ dt }\]

Answer 10

Yes, I assume you meant dy/dt. The problem is:
\[ y=\tan \theta \rightarrow dy/dt = \sec^2 \theta \frac{ d \theta }{ dt }\]

Answer 11

but sec^2 theta = cosx/sinx is it not?

Answer 12

No ...\[\sec^2 \theta=\frac{ 1 }{ \cos^2 \theta }\]

Answer 13

hmmm for some reason my lecture notes say that they are equal. I will have to make a correction than. But than tan theta = cosx/sinx? just like tan=y/x ?

Answer 14

I am sure of that as my professor clearly showed me an example where tan x is = sin x / cos x, but that is different to cosx/ sinx.....

Answer 15

Yes tan(x)=sin(x)/cos(x) is correct but its derivative is sec^2(x).

Answer 16

Whoora we got to the bottom of it. Thank you for your patience and diligence. This has helped me quite a lot and showed me some errors in my notes.

Answer 17

OK, you are very welcome!