# Thread: Antiderivative problems!!

1. ## Antiderivative problems!!

1. Find the particular antiderivative that satisfies the following conditions:

??

2. Find the particular antiderivative that satisfies the following conditions:

???

3. A particle is moving as given by the data below:

where is the position, and is the velocity.
????

2. Originally Posted by Kayla_N
1. Find the particular antiderivative that satisfies the following conditions:

??

2. Find the particular antiderivative that satisfies the following conditions:

???

3. A particle is moving as given by the data below:

where is the position, and is the velocity.
????
They are all done the same ways so here is the first

Since we know it is a derivative we can integrate both sides to get

$\int dR=\int \frac{40}{t^2}dt \iff R=-\frac{40}{t}+C$

Now we use the inital condition

$R(1)=40=-\frac{40}{1}+C \iff 80=C$

$R(t)=-\frac{40}{t}+80$

3. Originally Posted by Kayla_N
1. Find the particular antiderivative that satisfies the following conditions:

??

2. Find the particular antiderivative that satisfies the following conditions:

???

3. A particle is moving as given by the data below:

where is the position, and is the velocity.
????

CAn Someone please lead me into number 2. Where do i start? I look at number 1 and still bit confuse. Please help

4. 1.

Integrate both sides.

The integral of dR/dt is R. The integral of 40/t^2 is $\frac{-40}{t} + C$. Now you know that R = -40/t + C And when t = 1, R = 40. So, it follows that C = 40 + 40 = 80. Therefore, the particular integral (or antiderivative) is R(t) = -40/t + 80.

2.

Same thing. Integrate both sides.

p'(x) is really dp/dx. Its integral is p. You can find out the integral of 40/x^3. Do NOT forget the +C (there are no limits), or else you won't get a 'particular' integral. You know that p(4) = 4. That means, when x = 4, p = 4 as well. Find C, then get the particular integral.

I hope that helps.

ILoveMaths07.

P.S. Give it a try, show how you worked it out, and we'll help you out.

5. so the integral i got was -(20/x^2)+C. Now p=4 means x=4.

C= 21/4 and final answer i got is -20/x^2 +21/4.

Thanks for helping me...

CAN SOMEONE HELP ME WITH PLEASEEEEEEEE

6. Originally Posted by Kayla_N
so the integral i got was -(20/x^2)+C. Now p=4 means x=4.

C= 21/4 and final answer i got is -20/x^2 +21/4.

Thanks for helping me...

CAN SOMEONE HELP ME WITH PLEASEEEEEEEE
Please .... less of the hysterical pleading.

Your value for C is wrong, I have no idea how you got it. Check your arithmetic.

Q3 is done like the others. Note that by definition $\frac{ds}{dt} = v$.

7. Originally Posted by mr fantastic
Please .... less of the hysterical pleading.

Your value for C is wrong, I have no idea how you got it. Check your arithmetic.

Q3 is done like the others. Note that by definition $\frac{ds}{dt} = v$.

Sorry but my C value is correct. Thanks for helping me on 3, i got (-3sin(t)-3cos(t))+3.

8. Originally Posted by Kayla_N
Sorry but my C value is correct. [snip]
You're right. My mistake.

### content

Click on a term to search for related topics.