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[Silencher]

Hello.

I'm not sure what I'm doing wrong. These are homework problems, but the values will change the next time I attempt them, so let me show what I did.

--
1. A^3+B^3 = 28. Find B' when A = 3 and B' = 5

First I tried to just find 'B'. : B^3 = 28-(3)^3 | b^3 = 28-27. | b^3 = 1 | b = 1

Then I did:
3A^2*A' + 3B^2*B' = 0
A' = (3B^2*B') / 3A^2
A' = (3(1)^2 * 5) / 3(3)^2
A' = 15/27 or .555 ...

But the correct answer is: -5/9. And I don't know where I went wrong. I think probably when I found what B equals?

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2. A 20 ft ladder leans against a vertical wall. If the bottom of the ladder is sliding along the pavement directlya way from the wall at 2ft/sec, how fast is the height of the top of the ladder changing when the lower end of the ladder is 5 feet from the wall?

I labelled the ladder as 'C'. The wall as 'a' and the ground/etc as 'b'. I already know what may have happened is I switched A and B, but let me go through this anyway, if that was the only error, great. If not, then I can find any other errors.

So I have
A = 5 <--wrong, but this is what I tried.
A' = ? <--main thing to solve
B = ?
B' = 2ft/sec

So I tried to find B with the pythagorean(sp) theorum.

5 + b^2 = 20
b^2 = 15
b = sqrt(15)

leaving me with:
A = 5 <--wrong, but this is what I tried.
A' = ? <--main thing to solve
B = sqrt(15)
B' = 2ft/sec
C/C' = 20 (doesn't change - ladder)

So then I try to solve:

2A*A' + 2B*B' = 0
2A*A' = 2B*B'
A' = (2B*B')/2A
Plugged in values:
A' = ((2(sqrt15))*2)/(2(5))
A' =-1.549193338 (needed to round to thousandth)
A' = -1.549

But the correct answer was:
A' = -0.516

Where did I go wrong in this one? Or was it just that I got A and B switched around?

--

 

[DrPhil]

Hello.

I'm not sure what I'm doing wrong. These are homework problems, but the values will change the next time I attempt them, so let me show what I did.

--
1. A^3+B^3 = 28. Find B' when A = 3 and B' = 5

First I tried to just find 'B'. : B^3 = 28-(3)^3 | b^3 = 28-27. | b^3 = 1 | b = 1

Then I did:
3A^2*A' + 3B^2*B' = 0
A' = (3B^2*B') / 3A^2....Negative
A' = (3(1)^2 * 5) / 3(3)^2
A' = 15/27 or .555 ......Reduce the fraction, don't convert to decimal

But the correct answer is: -5/9. And I don't know where I went wrong. I think probably when I found what B equals?

B=1 when A=3 is correct. Your error was not to change the sign when you transposed 3B^2*B' to the right side

2. A 20 ft ladder leans against a vertical wall. If the bottom of the ladder is sliding along the pavement directlya way from the wall at 2ft/sec, how fast is the height of the top of the ladder changing when the lower end of the ladder is 5 feet from the wall?

I labelled the ladder as 'C'. The wall as 'a' and the ground/etc as 'b'. I already know what may have happened is I switched A and B, but let me go through this anyway, if that was the only error, great. If not, then I can find any other errors.

So I have

Let A = vertical position of top of ladder against wall
A' = ? <--main thing to solve
B = position of bottom of ladder away from the wall = 5 ft
B' = 2ft/sec

So I tried to find B with the Pythagorean theorem.

5^2 + A^2 = 20^2.......C^2 = A^2 + B^2
b^2 = 15......................A^2 = 375
b = sqrt(15)..................A = sqrt(375)

One place you messed up was not squaring the numbers in the Pythagorean theorem. And of course confusing A and B beyond any hope of recovery. Try again!

 

[Silencher]

Thanks! I retried it with the errors I made in mind and fixed them, thank you very much.