I am having issues graphing the first (Velocity) and second (Acceleration) derivatives of a function given to me in graph form.

enter image description here

(Sorry for the poor graph drawing)

I have a few questions:

  1. Is my graph for the first and second derivative correct and if not how can I fix it?
  2. How do I know where the graph is speeding up or slowing down?
  3. How is the graph speeding up at (1,2) U (3,4)?
  • $\begingroup$ Speeding up means the acceleration is positive, so the second derivative of the position function is positive. Have you learnt about how to inspect from a graph where the second derivative is positive? (For slowing down, everything is the same but replace "positive" with "negative".) This is based on the "concavity" or curvature of the graph, see for example this page for further reading: tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx. $\endgroup$ – Minus One-Twelfth Mar 7 at 22:41
  • $\begingroup$ If what I have is right then how is the graph speeding up at (1,2) $\endgroup$ – Aeryes Mar 7 at 22:42
  • $\begingroup$ @MinusOne-Twelfth Yes I have learned that some time ago. If the second derivative is below the x axis, that doesnt matter correct? As long as the slope is positive? My issue here is I dont see how the graph is speeding up at (1,2) $\endgroup$ – Aeryes Mar 7 at 22:44
  • $\begingroup$ @Aeryes Basically just by inspecting the position function graph (without needing to try to draw the second derivative graph, which can be a big tricky), if is possible to see where the graph has positive or negative second derivative. This is based on concavity, and the link in my last comment may be helpful, especially the pictures there. $\endgroup$ – Minus One-Twelfth Mar 7 at 22:48
  • $\begingroup$ @MinusOne-Twelfth But in this case, the velocity graph and acceleration graph have different signs which means they are slowing down? The answer however says that they are speeding up. $\endgroup$ – Aeryes Mar 7 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.