2 related rates problems, any direction would be appreciated

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xtreme1

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So I have two related rates problem that I have found troublesome. Can anyone tell me if my thinking on these is correct, I will post as much info on them as possible, so if you see a mistake please inform me.

Problem 1: A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon rises vertically upward at a rate of 10 feet per second, how fast is the distance from the observer to the balloon increasing when the balloon is 50 feet high?

This is my given info stated and what I am to find:
Given: Balloon is 150 feet away from observer,
dx/dt = 10 ft/s
Find: dy/dt when x=50

Here is my work:

Using Pythagorean Theorem: ((y)^2)=((x)^2) +((150)^2)

I then take the derivative of both sides and make sure the dy/dt is on its own side so I can solve for it: (dy/dt)= (x/y)* (dx/dt)

I then attempt to find out what y is since I already know what x and dx/dt are:
(y^2)=(x^2)+(150^2)
Sqrt(y^2)= Sqrt((10^2)+(150^2))
Sqrt(y^2)= Sqrt(25000)

I then put y into the derivative equation:
(dy/dt)= (50/Sqrt(25000)*(10)

Answer:
(dy/dt)= 3.1623 ft/s

Problem 2: Water stored in an inverted right circular cone leaks out at a constant rate of 2 cubic inches per day. The height of the tank is 100 inches and the radius of the base of the cone shaped tank is 25 inches. At what rate is the depth of the water inside the tank decreasing at the moment the water level is 40 inches deep.

This is my given info stated and what I am to find:
Given: Water leaks out of an inverted cone at 2 ((in)^3)/day
Height of tank: h=100
V=(1/3)(PI)(r^2)(h)
I needed to get r in terms of h so by similar triangles we know (r/25)=(h/100) or (r)=(1/4)(h)
Find: dv/dt when h=40 (I was a little uncertain whether I wanted to find dh/dt or dv/dt)

Here is my work: V= (1/3)*(PI)*(((1/4)(h))^2)*(h)
Simplified: V= ((PI)*( h^3))/(48)
Derivative: dv/dt= (PI)* (1/16)*(h^2)*(dh/dt)
I am really uncertain from here:
(dv/dt)=(PI)*(1/16)* ((40)^2)*(dh/dt)
(Dv/dt)=(PI)*(1/16)* (1600)*(-2)
Answer:
Dv/dt=(-200)(pi)? Or 628.3185 ((in)^3)/day

If you see any errors please let me know what they are and how I can fix them. I really appreciate the help.
 
Re: 2 related rates problems, any direction would be appreci

xtreme1 said:
Problem 1: A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon rises vertically upward at a rate of 10 feet per second, how fast is the distance from the observer to the balloon increasing when the balloon is 50 feet high?

This is my given info stated and what I am to find:
Given: Balloon is 150 feet away from observer,
dx/dt = 10 ft/s
Find: dy/dt when x=50
Click to expand...

You seem a bit confused on the movement of the balloon. "Upward" would suggest dy/dt = 10 fps.

Give it another go.
 
Re: 2 related rates problems, any direction would be appreci

Aren't x and y interchangeable as long as i keep them consistent, its not like its y on a graph moving upward, its x on a triangle with y standing for the hypotenuse
 
Re: 2 related rates problems, any direction would be appreci

Maybe, but not if you define them badly.

#1 - State your defintions CLEARLY. Write them down!

You did not do this in your initial presentation. This led directly to confusion.

#2 - How did you do this?

(y^2)=(x^2)+(150^2)
Sqrt(y^2)= Sqrt((10^2)+(150^2))

How did x get to be 10? You must have meant 50.

#3 - How did you do this?

Sqrt(y^2)= Sqrt((10^2)+(150^2))
Sqrt(y^2)= Sqrt(25000)

How did (10^2)+(150^2) get to be 25000? It's magically back to where it was before the error in the previous step.

Be careful. Be clear. Say what you are doing. Make sure your work can be followed.

Here's a model.

x = vertical distance travelled by the balloon.

Observe right triangle Observer - Balloon Starting Location - Current location of balloon. Call it Triangle OSC.

Y = Hypotenuse of Triangle OSC - The direct path from the observer to the current location of the balloon.

We are given side OS = 150 ft - The distance from the observer to the balloon starting point.

1) From the Pythagorean Theorem, we have: Y^2 = x^2 + 150^2

Given x = x(t) and Y = Y(t), we have the derivative with respect to 't'.

2Y(dY/dt) = 2x(dx/dt)

Solving for dY/dt gives: dY/dt = (x/Y)(dx/dt)

Given x = 50 ft
Using equation 1) to solve for Y gives: Y^2 = 50^2 + 150^2 = 2500 + 22500 = 25000 and Y = 50 * sqrt(10) ft

Given dx/dt = 10 fps

We have the final solution: dY/dt = (x/Y)(dx/dt) = (50/[50 * sqrt(10)]) * 10 fps = sqrt(10) fps

Be careful. Be clear. Say what you are doing. Make sure your work can be followed.

Note: You could improve on this presentation by being a little more formal with the units.
 
Re: 2 related rates problems, any direction would be appreci

For the second one, they want dh/dt. You got a little mixed up there.

\(\displaystyle \frac{dV}{dt}=\frac{{\pi}h^{2}}{16}\cdot \frac{dh}{dt}\)

Plug in dV/dt and h, then solve for dh/dt.
 
Re: 2 related rates problems, any direction would be appreci

okay thank you would dv/dt on the second be -2 and im solving for dh/dt?
 
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