Need help with Calc Questions #59.
Calculus

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First off, rewrite your units for answers 13.
5. Write a sign chart for the velocity function. The object "changing direction" means its velocity changes sign from  to + or from + to . The object has zero velocity at t = 0, 6 seconds, but the velocity is only changing from signs, from negative to positive, at t = 6 seconds.
6. Refer to sign chart of velocity function. The object's motion to the right means its velocity is positive. Find the interval of t for which the object has a positive velocity, which is all t > 6 seconds.
7. Write a sign chart for the acceleration function directly below the velocity sign chart. When something is "speeding up," its accelerating in the same direction that it is moving. When something is "slowing down," its acceleration opposes its motion. In the math, when velocity and acceleration have the same sign, the object is "speeding up;" when they have opposite sign, the object is "slowing down." With the velocity and acceleration sign charts drawn so that they share vertical line xvalues, their signs for any value of t can be easily compared.
8. Graphing the position function. First, find some simple values for position at a few inputs for t to show where the curve passes through. Then show when the position function isn't changing at t = 0, 6 at (0,3) and (6, 429). Then simply connect the points, keeping in mind the sign of the acceleration, which, for the graph of x, is concavity.
Please let me know if you need any clarification. I'm always happy to answer your questions.9. Assuming this problem is standing alone, draw the function f(x) = [tan(x)]/x. This graph is unique for a couple reasons, most notably its xaxis symmetry and that it has x in the denominator but is still defined at x=0. Plug in pi/4 to find the point the line must pass through; (pi/4, 4/pi). Use the quotient rule to derive, yielding f'(x) = [xsec^2(x)  tan(x)]/x^2. Plug in x = pi/4 to get the slope m of the tangent line to the curve at that input, which is m = (8pi  16)/pi^2. Lastly, put the slope m and the x and y components of the point (x = pi/4, y = 4/pi) into the pointslope formula for a line to get the desired equation for the line.
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