Topic awaiting preservation: Math Problems

I have some problems with math that my teacher couldn't answer. I tried answering them several times but still couldn't get the right answer for them.
Here they goes...

1. An oil spill spreads in a circular pattern on Moreton Bay. The area of the spill is increasing at a rate of 1 squaremetres/second. Find the rate of increase of the radius when the area is 50 square mteres.

2. Crusher dust is tipped onto a conical pile whose height remains equal to its radius. If the material is tipped at a rate of 2 cubicmetres/minute, at what rate is the radius increasing 10 minutes after the pile is started?

3. A water trough is 2 metres long and the cross-section is a right-angled isosceles triangle. Water enters the trough at a rate of 18 Litres/minute. At what rate is the water rising when the depth is 15 cm?


Thank you for helping!

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Deviations | 5464 | My Poetry Cell | My Own Domain | Support and advice needed. Now!
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| "The past will always attack the present with the pain of your memories." - Seiichi Kirima |
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I'll solve the first one for you, and you can use the same method for 2 and 3.

Area of a circle is Pi x R² and you know that dA/dT = 1, R = (SQR(50)/Pi). You're looking for dR/dT.

Find the derivative of the area of a circle:

dA/dT = 2 x Pi x R x dR/dT

So...

1 = 2 x Pi x (SQR(50)/Pi) x dR/Dt

dR/dT = 1/2(SQR(50)/Pi) = the change in radius over time when the area is 50M²

It's been a long time since I did my last calculus class, if I did something stupid someone point it out - I did this with no pen/paper nearby.
But assuming this is the right method, use the same method for the other two, neither of them seem particularly difficult.

Ooh, related rates. I'm assuming this is for calculus.

The general way to solve these problems is to take these steps:

1. take the information you have and write out an equation (or more than one) which describes what's going on
2. differentiate both sides of the equation(s) with respect to time
3. plug in the rates and values that you know, and solve for the ones that you don't

Remember these steps! They give you a method to get through a problem which is otherwise very daunting.

So, number one:

Oil in a circular pattern. The problem deals with area, so we'll need the function for the area of a circle:

A = pi*r^2

Now we differentiate with respect to time; note that area changes over time, and radius changes over time, so we have to treat them as though they're *functions* of t (A(t) and r(t)) instead of constant variables.

dA/dt = pi * 2*r * dr/dt

At this point we want to plug in the values we know so we can solve for the ones we don't. What we're ultimately trying to solve for is dr/dt, so we're going to need to know dA/dt and r, which are the other two unknowns in our equation. We're interested in what happens when the area is 50 square meters, so let's start by taking that fact and finding out what the radius is at that time (just by using the original equation):

A = pi*r^2
50 = pi*r^2
sqrt(50/pi) = r

Now, we know that the area is increasing at a rate of 1 squaremeter/second, so we can plug in 1 for dA/dt. We also know r at the time we're interested in. By plugging these into our differentiated equation, we can solve for dr/dt, the rate of increase of the radius:

dA/dt = pi * 2*r * dr/dt
1 = pi * 2*sqrt(50/pi) * dr/dt
1 / (pi * 2*sqrt(50/pi)) = dr/dt

So the answer is 1 / (pi * 2*sqrt(50/pi)) meters/second. You could calculate this and give it in decimal format or whatever. Also, be sure to check if the answer makes *sense*, in case you did something wrong; for instance, if you got a negative number, that wouldn't make sense because we know the radius is growing. Or if you got 2 million, there would be something wrong with that because we know it's not growing that fast.

_________________________________

Number 2:

This equation gives us information about the change in volume, and asks us about the radius. So we want to have an equation which involves the volume and radius of the cone. We know the equation for the volume of a cone in terms of its radius and height:

V = 1/3 * pi * r^2 * h

And we happen to know that the height is equal to the radius, so we can just replace h with r:

V = 1/3 * pi * r^3

There's our equation. As always, we differentiate with respect to time:

dV/dt = 1/3 * pi * 3*r^2 * dr/dt
dV/dt = pi * r^2 * dr/dt

Now, what we want to solve for is dr/dt (because the problem asked us what rate the radius is changing). We'll need to know dV/dt and r for this. dV/dt is given to us: 2 cubicmeters/minute. The radius is not given to us, we'll have to find it from the other piece of information we have: "10 minutes after the pile is started."

In order to figure out the radius at this point in time, let's look at what we know about the system. We know that the volume is changing at a rate of 2 cubicmeters/minute. This means that after 10 minutes, the volume is going to be 2 * 10 = 20 cubicmeters. We can plug this into our original equation to solve for r:

V = 1/3 * pi * r^3
20 = 1/3 * pi * r^3
(60/pi)^(1/3) = r

(How did I know to do this? Well, I knew that we had an equation relating radius to volume, and I knew something that related volume to time. So I made use of those two facts to get from time to volume to radius.)

Now we have everything we need to plug in and solve for the variable we want to know (dr/dt):

dV/dt = pi * r^2 * dr/dt
2 = pi * ((60/pi)^(1/3))^2 * dr/dt
2 / (pi * (60/pi)^(2/3)) = dr/dt

Our final answer is 2 / (pi * (60/pi)^(2/3)) meters/minute. Again, check the answer against common sense after you've calculated it.

_________________________

Now see if you can do number 3 on your own; it's really the same thing as the first two, just with a different equation because you have a different shape.


 

I feel so stupid, thank you.

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Deviations | 5464 | My Poetry Cell | My Own Domain | Support and advice needed. Now!
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| "The past will always attack the present with the pain of your memories." - Seiichi Kirima |
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(Edited by Yannah on 05-30-2005 09:38)

Don't feel stupid; these aren't easy problems, especially when you're new to them. Let us know if you get stuck on the third one or any others.


 

Oh thank you so much... now I feel so relieved. You are like the brothers I've never had.
*Ran towards Slime and Dan for a group hug*

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Deviations | 5464 | My Poetry Cell | My Own Domain | Support and advice needed. Now!
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| "The past will always attack the present with the pain of your memories." - Seiichi Kirima |
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(Edited by Yannah on 05-30-2005 11:22)

3.)
Doing it in cm/min
h=15 cm
Length=2 m= 200cm
Sin 45=h/x
x=h/sin45
Area= 1/2x^2
Volume= 1/2x^2*200=100x^2
Volume= 100*(h/sin45)=200h^2
Therefore, dV/dh=400h
We're trying to look for dh/dt
dV/dt=(dV/dh)*(dh/dt)
dV/dt=400h*(dh/dt)
18L/400h=dh/dt [1000cm^3=1L]
18 000/(400*15)=dh/dt
dh/dt=3cm/min is the rate of the water rising when the depth is 15 cm.

*Did I do it right?

I got another problem.

A water trough is a prism with a parabolic cross-section. The trough is 3m long and the cross-section corresponds to the part of y=(5x^2-400x+350)/90 that is below the x-axis, where the units are in centimetres. Find the area of the cross-section and the volume of water that can be contained in the trough.
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Deviations | 5464 | My Poetry Cell | My Own Domain | Support and advice needed. Now!
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| "The past will always attack the present with the pain of your memories." - Seiichi Kirima |
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(Edited by Yannah on 05-31-2005 05:40)