Need help with Calc Questions #5-9.

Calculus
Tutor: None Selected Time limit: 1 Day

Sep 24th, 2015

Thank you for the opportunity to help you with your question!

First off, rewrite your units for answers 1-3. 

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 x-values, 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.
Sep 28th, 2015

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 x-axis 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 point-slope formula for a line to get the desired equation for the line.

Sep 28th, 2015

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Sep 24th, 2015
...
Sep 24th, 2015
Dec 4th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer