# Calculus Continuous Function Problem

## Homework Statement

Find A and B so that f(t) is continuous everywhere.

## Homework Equations

Suppose that:
[PLAIN]http://img690.imageshack.us/img690/8531/eqwkshp3.png [Broken]

## The Attempt at a Solution

Well, I wouldn't be posting if I wasn't lost, but I will tell you what I've tried. I know that I have to set the middle equation equal to either the first or third and solve for A or B. However, if I set the middle equation equal to the first one and then plug in t = 0, it ends up canceling out the entire one side. So then I've tried setting the middle equation equal to the third equation and plugging in t = 5 to solve for B and I end up with B = (46 - 6sin(5A))/5. I'm probably doing something incorrectly.

I know that for these types of problems you solve for the one variable and then you plug it back in to solve for the other one, but I don't think I'm doing this particular problem correctly. Any guidance would be appreciated. Thanks.

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Dick
Homework Helper
You can't just plug in for the second equation. You need to take the limit as t goes to zero of sin(At)/t. Otherwise, you get 0/0 which doesn't mean anything. Any ideas on that limit?

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You can't just plug in for the second equation. You need to take the limit as t goes to zero of sin(At)/t. Otherwise, you get 0/0 which doesn't mean anything. Any ideas on that limit?
I know that if it was just sin(t)/t that the limit would be 1 (for this equation A would have to be 1 to make that work).

Dick
Homework Helper
I know that if it was just sin(t)/t that the limit would be 1 (for this equation A would have to be 1 to make that work).
That's a good start. Do you think if limit t->0 of sin(t)/t=1 then limit t->0 sin(At)/(At) might also be 1? If so then what's limit t->0 of sin(At)/t?

That's a good start. Do you think if limit t->0 of sin(t)/t=1 then limit t->0 sin(At)/(At) might also be 1? If so then what's limit t->0 of sin(At)/t?
I think I have it:

lim t-> 0 sin(At)/t

1. Multiply by A to get: lim t->0 Asin(At)/At
2. Put the A out front: A lim t-> 0 sin(At)/At
3. lim t-> 0 sin(At)/At = 1 and then A * 1 = A

So the lim t->0 sin(At)/t = A?

Dick
Homework Helper
I think I have it:

lim t-> 0 sin(At)/t

1. Multiply by A to get: lim t->0 Asin(At)/At
2. Put the A out front: A lim t-> 0 sin(At)/At
3. lim t-> 0 sin(At)/At = 1 and then A * 1 = A

So the lim t->0 sin(At)/t = A?
Yes, you've got it.

Now, I'm assuming that the equation is now:

When t = 0

30(sin(At)/t) + Bt = 3∏
30 lim t->0 (sin(At)/t) + B(0) = 3∏
30A + 0 = 3∏
A = 3∏/30

Dick
Homework Helper
Now, I'm assuming that the equation is now:

When t = 0

30(sin(At)/t) + Bt = 3∏
30 lim t->0 (sin(At)/t) + B(0) = 3∏
30A + 0 = 3∏
A = 3∏/30
Looks good so far.

Looks good so far.
When t = 5

30(sin(At)/t) + Bt = -10[log(t)/log(5^(1/3))] + 76
30(sin(0.314*5)/5) + B(5) = -10[log(5)/log(5^(1/3))] + 76
0.164 + 5B = 46
5B = 45.84
B = 9.168

Dick
Homework Helper
Try the sin part again. You actually don't have to approximate pi. It works out to a nice number exactly.

Try the sin part again. You actually don't have to approximate pi. It works out to a nice number exactly.
And so it does indeed :P

30(sin(At)/t) + Bt = -10[log(t)/log(5^(1/3))] + 76
30(sin(3∏/30*5)/5) + B(5) = -10[log(5)/log(5^(1/3))] + 76
30(1/5) + 5B = 46
6 + 5B = 46
B = 8

Thank you so much for your help. =D