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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,411

Some problems...one at a time.

Suppose the integral was already performed on an equation. What is the simplified answer if you now take the integral from 0 to 1 for the following integrated equation:

8x^3 + 3x^2 - 4x + C

I got 7.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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7 is correct.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**ShivamS****Member**- Registered: 2011-02-07
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Consider 'x' to be a variable. What is the derivative of 12x^4 + 10x^3 - 5x^2 + 16 with respect to x?

I get 48x^3 + 30x^2 - 10x

Also:

Consider 'x' to be a variable. What is the integral of 18x^2 - 10x + 3 with respect to x? Pick the most correct solution.

I get 6x^3 - 5x^2 + 3x + C

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Both are correct! Good work!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,411

Thank-you; I just thought I needed some extra practive before the exams. After that, I should be much more relaxed... Anyways, I will post some more later.

*Last edited by ShivamS (2014-03-09 04:12:35)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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You are welcome.

Okay, post the other questions when you want.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,411

Hi; few other simple problems. Fred is standing on the ground and throws a ball up in to the air. He observes that it falls back down to the ground after 5 seconds. What was the initial velocity of the ball? I get 0 = -400 + 5V subscript 0 Thus initial velocity = 80ft / s

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,411

One more: Junita is standing on the roof of a building 192 feet tall and throws a ball up in to the air with an initial speed of 64 feet / second.

(a) Find formula for velocity and position of ball at a later time. I get v(t) = -32t + 64 And for position: y(t) = -16t^2 + 64t +192

(b) How high does the ball go? I am getting 256 feet,

(c) When does the ball hit the ground? I think this is probably wrong, but I am getting 6 seconds.

Thank-you in advance.

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**anonimnystefy****Real Member**- From: The Foundation
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I am getting 25 m/s. Which formula are you using?

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**ShivamS****Member**- Registered: 2011-02-07
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Hi Stefy; for which question are you getting that?

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**anonimnystefy****Real Member**- From: The Foundation
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The one in post #7.

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**ShivamS****Member**- Registered: 2011-02-07
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I get y(t) = Integral (-32t + v subscript 0)dt = -16t^2 + v subscript 0 t + C

So y(t) = -16t^2 + v0t (o = subscript)

So 0 = -16 * 5^2 + initial velocity * 5 = -400 + 5 (initial velocity) = -400 + 5 initial velocity

Solving gives us V0 = initial velocity = 80 ft / second

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**anonimnystefy****Real Member**- From: The Foundation
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Shivamcoder3013 wrote:

y(t) = Integral (-32t + v subscript 0)dt

How'd you get this?

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**ShivamS****Member**- Registered: 2011-02-07
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Similar to the problem in post 8. ground level will be position 0, up will be the positive direction and time t=0 will be the instant when Fred first throws the ball. The acceleration due to gravity will be a = -32. Thus dv/dt = -32, so that v(t) = -32 + C. The problem does not give any information about the velocity of the ball at any time so I use v0 to represent initial velocity. AT time t = 0, I get v(t) = -32t + v0. So, as v(t) = dr/dt, y(t) = integral (-32t + v0)dt = -16t + v0t + C.

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**anonimnystefy****Real Member**- From: The Foundation
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But, the acceleration due to gravity is 9.81 (or 10, if you round it up) mps.

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**ShivamS****Member**- Registered: 2011-02-07
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Well, the units in the problem were feet, and 9.81 is approximately 32 feet. Also, "Similar to the problem in post 8. ground level will be position 0, up will be the positive direction and time t=0 will be the instant when Fred first throws the ball. The acceleration due to gravity will be a = -32."

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**anonimnystefy****Real Member**- From: The Foundation
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Ah, so we are both correct on the initial velocity for the one in post #7.

Let me look at the second one.

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**ShivamS****Member**- Registered: 2011-02-07
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25m is 82 feet so did you simply approximate something?

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**anonimnystefy****Real Member**- From: The Foundation
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Yeah, I took g to be 10m per second squared. I am getting approximately 24.525mps with g=9.81m per second squared.

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**ShivamS****Member**- Registered: 2011-02-07
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Oh, okay. Thank-you. The second one is a bit more complicated.

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**anonimnystefy****Real Member**- From: The Foundation
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I think all three are correct.

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**ShivamS****Member**- Registered: 2011-02-07
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Okay, thank you very much. I have some more but I will post them later. Actually, here are some vector calculus questions:

Determine if the vector ﬁeld f(x, y,z)= x yzi+xzj+x yk has a potential in R3

I am getting that it does not have the potential.

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**anonimnystefy****Real Member**- From: The Foundation
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I don't think I know what the potential of a vector is.

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**ShivamS****Member**- Registered: 2011-02-07
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http://en.wikipedia.org/wiki/Vector_potential

But thank-you anyways.

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**ShivamS****Member**- Registered: 2011-02-07
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Thanks a lot Stefy; the midterm exam went extremely well.

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