1. ## related rates problems..

1. Water flows into a vertical cylindrical tank at the rate of 12 cu.ft/min. The surface rises 6 inches per minute. Find the radius of the tank.

2. If the angle of elevation of the sun is 45degrees and is decreasing at 0.25rad/hour, how fast is the shadow cast on level ground changing(lengthening) when the pole is 50ft tall?

My Effort..

1.

Sol'n:
V=Bh
V=pi (r^2) h --> is used this as my working equation

dV/dt = 2pir(dr/dt)(dh/dt)

..i stopped in here because it noticed that there is no rate for radius, so what I did is to work again..

V=Bh
V=pi (r^2)h
r= sqrt.(V/pi h) --> this is my another working equation

r = sqrt[ (dV/dt)/ pi (dh/dt) ]
r = 2.76 ft

There it goes.. I don't know if it is correct.

2.

Sol'n:

tan 45 = 50/x
x = 50/tan45 --> this serves as my working equation

dx/dt = [-50sec^2 (45) (0.25rad/hr)] / tan^2 (45)
dx/dt = -25ft/hr

There is all my effort. Tell me all the corrections.

Thanks!

2. 1. $r$ is a constant

$\frac{dV}{dt} = \pi r^2 \cdot \frac{dh}{dt}$

sub in the given values and solve for $r$

2. $\theta = \arctan\left(\frac{50}{x}\right)$

take the time derivative of the above equation, sub in the given value for $\frac{d\theta}{dt}$ and solve for $\frac{dx}{dt}$

when $\theta = 45^\circ$ , $x = 50$ ... also remember that $\frac{d\theta}{dt} < 0$

3. so in short Sir, all my efforts are correct or wrong? )

4. Originally Posted by cutiemike1
so in short Sir, all my efforts are correct or wrong?
1. the value you arrived at for the radius is correct ... but you sure took the long road to get there.

2. incorrect

5. Originally Posted by skeeter
2. $\theta = \arctan\left(\frac{50}{x}\right)$
If i use this as my working equation, i will not have a value for 'x' since x is not given.

6. Originally Posted by skeeter
when $\theta = 45^\circ$ , $x = 50$ ...
did you not read this part of my first response?

7. hmmm...

sir, i tried both equations inverse and not inverse, it just came up with the same answer..

this is how i made them both..try to see if there is a mistake,

the second column is to how I understand your explanation awhile ago.

8. I note that you corrected one error ... replacing 45 in your initial set up with $\theta$.

also, note that $\frac{dx}{dt} = 25 \, ft/hr$ ... positive, not negative. can you determine why? (I told you to remember something in my earlier post)

9. Yeah! I remember now..since the elevation of the sun is decreasing the angle should be negative..am i right?

Thanks a lot for the help! ^_^ I really appreciate the whole thing.

By the way, it is our examination week this week, as for our first schedule today, calculus, can you give me some tips? The coverage of the exam are differentiation, related rates, tangent line and normal line and applications of derivatives of trigonometric and inverse trigonometric function.

Thanks a lot!

10. Originally Posted by cutiemike1
Yeah! I remember now..since the elevation of the sun is decreasing the angle should be negative..am i right?
correction ... $\frac{d\theta}{dt}$ is negative because $\theta$ is decreasing

11. Many thanks to you Sir Skeeter... ^_^

just wanted to share...I did great somehow in our exam in calculus a while ago ... ^_^

THANK YOU!!