<p>a particle is moving to the right.
x(t)=cos(pi*t^2) 0 less than or equal to t less than or equal to 3 </p>
<p>what is the velocity as a function of t?
what is the acceleration as a function of t?
what are the values when it is moving right?
what is the acceleration at an instant when it returns to t?</p>
<p>I'm going to let u=(pi t^2) so I don't have to write it out</p>
<p>The velocity is going to be x'(t), so it's x'(t)= -sin (u) (u')
Acceleration is x''(t)= -cos(u)(u')^2 - sin(u)(u')
Set x'=0 and when the value is positive, the particle is moving right
I am not sure what the fourth part is asking.</p>
<p>How about this one?</p>
<p>You are making a box from 24 by 26cm of cardboard by cutting congruent squares from the corners.</p>
<p>What are the maximum dimensions you could get?
What is the volume?</p>
<p>Wouldn't maximum dimensions and maximum volume be the same? </p>
<p>So basically, you have a rectangle with four squares drawn in the corners. Each side of the square will be x. So the length of the rectangle is now 26-2x (original length minus two sides of a square) and width is 24-2x. The forumla for volume is LxWxH as I'm sure you know, so since the height is x (if you cut out the squares and fold the sides up, you'll see the height is x), the formula is (26-2x)(24-2x)(x)= Volume. Take the derivative of that and solve it for zero, plugging in the x value for each of the expressions for the width, length, and height. I presume they mean find the dimensions for the largest box, and then multiply them to get the volume.</p>
<p>I think what he meant by the maximum dimensions was the maximum surface area of the box you can get from that cardboard (after its corners are cut). That would be to maximize the function f(x) = (26-x)(24-x) + 2(26-x)x + 2(24-x)x.</p>
<p>So should I find the deriv of f(x) = (26-x)(24-x) + 2(26-x)x + 2(24-x)x to help me find my answer?</p>
<p>Take the derivative of that that function, then find the critical points. Do the second derivative test to see if it's a max or min. However, that one is only for finding the maximum dimensions. The x for which the volume is maximum can be gotten by deriving the volume function Patrick wrote above.</p>