Calculus Optimization/Related Rates. Update: Took test, got it back.

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iluvtruenos

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Apr 14, 2005
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His exact words were:

Test on Friday on optimization and related rates. See you then.

It's Wednesday, and I'm barely grasping related rates. Can anyone point me in the right direction regarding them? I'm really, really screwed.

UPDATE: Took test, and yep. Failed. 41%. Motherfukcer.
 
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hypn0tik

Diamond Member
Jul 5, 2005
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Perhaps post a problem that you're having difficulty with? It might be easier to help out.

Edit: Well, as far as starting off in the right direction goes, DRAW A PICTURE! I can't stress how important that is.
 
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RadioHead84

Platinum Member
Jan 8, 2004
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OK here ya go..I had a test on this a few months ago. Basically it would be helpful if you gave us an examle question and we could work it through with you.
 
pray4mojo

pray4mojo

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Mar 8, 2003
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usually when i don't know what to do i take the derivative then set it equal to 0.
 
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RadioHead84

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Jan 8, 2004
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Bascially you need to find the equation for the problem. Then take the derivative of the equation BUT instead of just taking the derivative with respect to x you have to do it for y too.

For example.

X^4 would be 4X^3 *Xprime(or the derivitive of x)

Same goes for Y.

Now in the problem there should be something saying the rate of change of these varieables..and since the deriviative is the rate of change plug and chug.

Really an example would help tho.
 
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iluvtruenos

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Apr 14, 2005
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Ok, yes, here's a problem.

Ship A is heading West at a rate of 15km/h towards a lighthouse.

Ship B is heading North at a rate of 10 km/h away from the lighthouse.

What is the change in distance between them when A is 4km away from the lighthouse, and B is 3?

I got 6 for that, by using pythagorean theorem, taking the derivative, and solving for dz/dt.
 
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RadioHead84

Platinum Member
Jan 8, 2004
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Ok draw a picture of the problem that helps the most.

Your right about using the pythagorea theorem.

List what you know.
X prime = 15
Y Prime = 10
x=4
y=3
z=5

x^2 + Y^2 = Z^2


2x*xprime + 2y*Yprime = 2z*zprime

Plug and chug
120+60/2=18

This could be wrong....but its in the right direction lol
 
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hypn0tik

Diamond Member
Jul 5, 2005
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Well, start by drawing a picture. A few hints:
1) If you can, try and set up te problem on cartesian coordinates.
2) Try to make things as symmetric as possible.
3) In examples such as the one you posted, put the things that don't move at the origin.
4) Place everything else where it needs to go.
5) Make sure your signs are right.
6) Solve.

So, in this case, put the lighthouse at the origin. The positive x-axis will be East.

Now, you are given that Ship A is heading towards the lighthouse heading West. Therefore, it must be in the positive x-axis coming towards the origin. For this reason, dx/dt must be negative.
Ship B must be on the positive y-axis moving away from the origin. Here, dy/dt is positive.

When Ship A is at (4,0) and Ship B is at (0,3), the distance between them is 5, given by:

z^2 = x^2 + y^2
Differentiating:

2z dz/dt = 2x dx/dt + 2y dy/dt

z = 5
x = 4
dx/dt = -15
y = 3
dy/dt = 10

Solve for dz/dt

I hope this helps.
 
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RadioHead84

Platinum Member
Jan 8, 2004
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Originally posted by: hypn0tik
Well, start by drawing a picture. A few hints:
1) If you can, try and set up te problem on cartesian coordinates.
2) Try to make things as symmetric as possible.
3) In examples such as the one you posted, put the things that don't move at the origin.
4) Place everything else where it needs to go.
5) Make sure your signs are right.
6) Solve.

So, in this case, put the lighthouse at the origin. The positive x-axis will be East.

Now, you are given that Ship A is heading towards the lighthouse heading West. Therefore, it must be in the positive x-axis coming towards the origin. For this reason, dx/dt must be negative.
Ship B must be on the positive y-axis moving away from the origin. Here, dy/dt is positive.

When Ship A is at (4,0) and Ship B is at (0,3), the distance between them is 5, given by:

z^2 = x^2 + y^2
Differentiating:

2z dz/dt = 2x dx/dt + 2y dy/dt

z = 5
x = 4
dx/dt = -15
y = 3
dy/dt = 10

Solve for dz/dt

I hope this helps.
Click to expand...


DOH i read that they were both going away...My bad...My idea was right just read to fast.
 
I

iluvtruenos

Banned
Apr 14, 2005
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Originally posted by: RadioHead84
Originally posted by: hypn0tik
Well, start by drawing a picture. A few hints:
1) If you can, try and set up te problem on cartesian coordinates.
2) Try to make things as symmetric as possible.
3) In examples such as the one you posted, put the things that don't move at the origin.
4) Place everything else where it needs to go.
5) Make sure your signs are right.
6) Solve.

So, in this case, put the lighthouse at the origin. The positive x-axis will be East.

Now, you are given that Ship A is heading towards the lighthouse heading West. Therefore, it must be in the positive x-axis coming towards the origin. For this reason, dx/dt must be negative.
Ship B must be on the positive y-axis moving away from the origin. Here, dy/dt is positive.

When Ship A is at (4,0) and Ship B is at (0,3), the distance between them is 5, given by:

z^2 = x^2 + y^2
Differentiating:

2z dz/dt = 2x dx/dt + 2y dy/dt

z = 5
x = 4
dx/dt = -15
y = 3
dy/dt = 10

Solve for dz/dt

I hope this helps.
Click to expand...


DOH i read that they were both going away...My bad...My idea was right just read to fast.
Click to expand...

Yes, that's right because you multiply 4-distance A is from the lighthouse times -15 - rate at which it's going to it, add to 3 - distance b from lh, times 10, rate.

Then you get 30, and divide that by z, or 5, and you get the rate to be 6km/hr for the change.

I am not sure whether I'm right or not, but I think that this is how you solve it.
 
JujuFish

JujuFish

Lifer
Feb 3, 2005
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Optimization generally involves formulating an equation, taking the derivative and setting it equal to zero, then in the original equation checking the values at the endpoints and where the derivative equals 0. Whichever is highest/lowest is the answer you seek.
 
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RadioHead84

Platinum Member
Jan 8, 2004
2,166
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Originally posted by: JujuFish
Optimization generally involves formulating an equation, taking the derivative and setting it equal to zero, then in the original equation checking the values at the endpoints and where the derivative equals 0. Whichever is highest/lowest is the answer you seek.
Click to expand...

Yeah thats optimization for ya.

Its more tricky then the rates problem in my opinion..In some cases if you cant find the end points you need to take the second derivative to use the second deriviative test.
 
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hypn0tik

Diamond Member
Jul 5, 2005
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Originally posted by: RadioHead84
Originally posted by: JujuFish
Optimization generally involves formulating an equation, taking the derivative and setting it equal to zero, then in the original equation checking the values at the endpoints and where the derivative equals 0. Whichever is highest/lowest is the answer you seek.
Click to expand...

Yeah thats optimization for ya.

Its more tricky then the rates problem in my opinion..In some cases if you cant find the end points you need to take the second derivative to use the second deriviative test.
Click to expand...

You should always use a second derivative test to see if the result you obtained is infact a maximum or minimum depending on what you're looking for.
 
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desteffy

Golden Member
Jul 16, 2004
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perhaps there is some sort of ... textbook... you could read on the topic


hehe good luck!
 
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