<p>^for doh's question, I would integrate v(t) = sin (13-(t^2)) between the intervals 0 and 3 (first making sure that the curve doesn't become negative--this one doesn't). The area under velocity graph gives distance traveled. The answer is 0.52, but since x(0)=5, you add 5 to 0.52 to give x=5.52. </p>
<p>Don't we need the y function to find the position in the form (x,y)??</p>
<p>Damn! i'm stumped on the easy ones. Help me (and explain)</p>
<p>If a function f is given by f(x)=x^3 has an average value of 9 on the interval [0,k] then k=</p>
<p>A. 3 B. 3^.5 C.18^(1/3) D. 36^(1/3) E. 36^(1/4).</p>
<p>Answer is E. I was almost certain it would be A. Now, im startin to panic</p>
<p>9 = 1/(k-0) Integral x^3 dx from 0 to k
9 = 1/(k) (k^4/4 - 0)
9 = 1/(k) (k^4/4)
9 = k^3/4
36 = k^3
k = 36^(1/3)</p>
<p>I get D?</p>
<p>ye thats right sorry, i wrote the answers in the wrong order (lol)</p>
<p>Answer to Yatta!'s problem. (4x^8+5)(2x) Is this right? Fundamental Theorem of Calculus?</p>
<p>I had a question on an inverse problem. The question was:</p>
<p>Let f be the function defined by f(x) = x^3 + x. If g(x) = f^-1(x) (f inverse of x, sorry i'm not really sure how to type it properly), and g(2) = 1, what is the value of g'(2)?</p>
<p>(A) 1/13
(B) 1/4
(C) 7/4
(D) 4
(E) 13</p>
<p>The answer is B. I have no idea why and was hoping someone could explain it to me. According to the test im looking at, only 46% of people who got a 5 got this question right</p>
<p>so you start with
g'(2) = 1 / f'(g(2))</p>
<p>but I don't know why.</p>
<p>This is really common because no one really remembers this theorm.
g'(x)= 1/(f'(g(x)). So you know that g'(2) = 1/ f'(g(2)) right? g(2) is given, so g'(2)= 1/f'(1)
You are given the function f(x), so lets take the derivative of that and plug in 1 wherever we see x. So we get 3(1)^2+1=4 so g'(2) = 1/4
Hope this helped</p>
<p>Any help with this one?</p>
<p>A cube is contracting so that it's surface area decreases at the constant rate of 72 in^2/sec. Determine how fast the volume is changing at the instant when the surface area is 54ft^2.</p>
<p>To corn_nutz:</p>
<p>Thanks so much for the explanation. I fully understand it now</p>
<p>I'm not too sure if this is right or not, but you need to find a relationship for Surface Area to Volume. Surface Area is 6s^2 and Volume is s^3
So you find that s^2 = (Surface Area)/6 Plug that into the equation for volume and you get Volume = ((Surface Area)/6)^(3/2)
You know that the change in surface area is 72 in^2/sec so that's your dA/dt
You are trying to find dV/dt
Find the derivative of that equation that we just synthesized.
dV/dt = 1.5((Surface Area)/6)^(1/2)(dA/dt)
You know dA/dt and you are trying to find the instant the surface area is 54ft^2 Plug and solve for dV/dt
I noticed that your rate is in inches and your surface area is in feet, so you would have to compensate by changing the units.</p>
<p>Amazing. So simple!
I'd been struggling with that dang problem all day. </p>
<p>Corn_nutz: THANK YOU THANK YOU THANK YOU.</p>
<p>help please with this one?</p>
<p>The velocity, in ft/sec, of a particle moving along the x-axis is given by the function v(t) = e^t + te^t. What is the average velocity of the particle from time t = 0 to time t = 3 ?</p>