# Two antiderivative problems!!!

• Apr 6th 2009, 05:42 PM
Kayla_N
Two antiderivative problems!!!
1. Evaluate the indefinite integral.

http://webwork1.math.utah.edu/webwor...7080cae3c1.png

2. A stone is thrown straight up from the edge of a roof, http://webwork1.math.utah.edu/webwor...f6cf2003b1.png feet above the ground, at a speed of http://webwork1.math.utah.edu/webwor...07f0fb5471.png feet per second.

A. Remembering that the acceleration due to gravity is http://webwork1.math.utah.edu/webwor...0b8bfeb8d1.png, how high is the stone http://webwork1.math.utah.edu/webwor...4118766791.png seconds later?

B. At what time does the stone hit the ground?

C. What is the velocity of the stone when it hits the ground?
• Apr 6th 2009, 06:01 PM
Reckoner
Quote:

Originally Posted by Kayla_N
1. Evaluate the indefinite integral.

Let $u=5+14x^4\Rightarrow du=56x^3\,dx.$

Quote:

2. A stone is thrown straight up from the edge of a roof, http://webwork1.math.utah.edu/webwor...f6cf2003b1.png feet above the ground, at a speed of http://webwork1.math.utah.edu/webwor...07f0fb5471.png feet per second.

A. Remembering that the acceleration due to gravity is http://webwork1.math.utah.edu/webwor...0b8bfeb8d1.png, how high is the stone http://webwork1.math.utah.edu/webwor...4118766791.png seconds later?
Integrate the acceleration function to get the velocity function $v.$ You can solve for the constant of integration by noting that $v(0)=20.$

Integrate $v$ to get the stone's position function $s.$ Set $s(0)=775$ and you can solve for the constant of integration. Then just evaluate $s(3).$

Quote:

B. At what time does the stone hit the ground?
That is, for what value of $t$ is $s(t)=0?$

Quote:

C. What is the velocity of the stone when it hits the ground?
Evaluate the velocity function at the time found in part (B).
• Apr 6th 2009, 09:23 PM
Kayla_N
I still dont understand what you wrote for number 2..can you explain a little bit more.???
• Apr 6th 2009, 09:32 PM
Reckoner
Quote:

Originally Posted by Kayla_N
I still dont understand what you wrote for number 2..can you explain a little bit more.???

Velocity is the derivative of position; acceleration is the derivative of velocity.

The stone is undergoing a constant $-32\text{ ft./s}^2$ acceleration due to gravity, so the acceleration function is

$a(t)=-32.$

The velocity function is therefore

$v(t)=\int a(t)\,dt=\int(-32)\,dt=-32t+C_0.$

We know that the initial velocity is 20 ft./s, so

$v(0)=20\Rightarrow -32\cdot0+C_0=20\Rightarrow C_0=20$

and

$v(t)=20-32t.$

Work similarly to find the position function. Can you continue?
• Apr 6th 2009, 09:53 PM
Kayla_N
Quote:

Originally Posted by Reckoner
Velocity is the derivative of position; acceleration is the derivative of velocity.

The stone is undergoing a constant $-32\text{ ft./s}^2$ acceleration due to gravity, so the acceleration function is

$a(t)=-32.$

The velocity function is therefore

$v(t)=\int a(t)\,dt=\int(-32)\,dt=-32t+C_0.$

We know that the initial velocity is 20 ft./s, so

$v(0)=20\Rightarrow -32\cdot0+C_0=20\Rightarrow C_0=20$

and

$v(t)=20-32t.$

Work similarly to find the position function. Can you continue?

since v(t)=20-32t, so t= .625.
Then to find s(t)= \int (-32+20)dt
s(t)=-16t^2+20t+C2
3=-16(0)+20(0)+c2
C2=3??
Then i dont know what else to do!!!
• Apr 6th 2009, 10:02 PM
Reckoner
Quote:

Originally Posted by Kayla_N
since v(t)=20-32t, so t= .625.

At $t=0.625,$ the velocity is zero (i.e., the stone is not moving). This will occur when the stone reaches its maximum height. Were you asked to find this?

Quote:

Then to find s(t)= \int (-32+20)dt
s(t)=-16t^2+20t+C2
3=-16(0)+20(0)+c2
What are you doing? The initial height is 775 ft., not 3. After you solve for $C_2,$ evaluate $s(3)$ to get the position after 3 seconds.
• Apr 6th 2009, 10:10 PM
Kayla_N
So the s(t) -16t^2+20t+739 right???
• Apr 6th 2009, 10:14 PM
Reckoner
Quote:

Originally Posted by Kayla_N
So the s(t) -16t^2+20t+739 right???

Where did you get 739 from?
• Apr 6th 2009, 10:17 PM
Kayla_N
Quote:

Originally Posted by Reckoner
Where did you get 739 from?

well 739 is wrong, i did again 775=-16t^2+20t+C
then i plug in t=3 and i got C=859?

Jeeze i'm so lost.
• Apr 6th 2009, 10:21 PM
Reckoner
Quote:

Originally Posted by Kayla_N
well 739 is wrong, i did again 775=-16t^2+20t+C
then i plug in t=3 and i got C=859?

Think about what you are doing. The stone's initial height (at 0 seconds) is 775 ft., so when $t=0,\;s(t)=775.$ So why are you setting $t=3?$
• Apr 6th 2009, 10:24 PM
Kayla_N
i'm trying to calculate the height at 3 seconds.. Can you please show me how to get part A??
• Apr 6th 2009, 10:29 PM
Reckoner
Quote:

Originally Posted by Kayla_N
i'm trying to calculate the height at 3 seconds.. Can you please show me how to get part A??

Do as I said:

We have $s(t)=-16t^2+20t+C_2,$ and $s(0)=775$ (as I mentioned above). Therefore,

$s(0)=-16\cdot0^2+20\cdot0+C_2=775$

$\Rightarrow C_2=775$

and

$s(t)=-16t^2+20t+775.$

Now find $s(3).$
• Apr 6th 2009, 10:35 PM
Kayla_N
ok and i got 691. From there i went on to second part...s(t)=0
0=-16t^2+20t+775. I used quadratic formala and got 7.6127 sec. then moved on to 3 part, i used V= (7.6127)(-32)+20= -223.6068.