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I have a calculus question which i will display here as an image: enter image description here

I am interested to understand part (b) of this question. I actually got the answer, but i feel i need more to understand how to determine the maximum just via second derivative. If one takes the function P(t) and graphs it, you can see the largest value of the derivative happen to be on the Maxima of the graph of the derivative. Now if one was to take the derivative of this graph then you would get that the maximum rate of change happens when the second derivative is zero, which means its at the Points of Inflection. So it seems that for Trig functions the maximum rate of change happens at the points of inflection. So this is how i analyzed it, but i feel there is a better way to explain this.

Hope to get further insight from others here on the forum.

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2 Answers 2

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When I think about the 2nd derivative, I imagine the tangent line to the curve at a point $x$ and let $x$ increase. The 2nd derivative tells you about the change in the slope of the tangent line. So if the 2nd derivative is positive, the slope is increasing and so the tangent line is rotating counter-clockwise (as $x$ increases.) Likewise a negative 2nd derivative shows that the tangent line is rotating clockwise.

An inflection point is a point where the tangent line crosses the graph and it's also a point where the rotation changes from clockwise to counter-clockwise, or vv.

So you get a maximum slope exactly when the rotation changes from CCW to CW. Which is exactly that the inflection points.

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  • $\begingroup$ So Goddard, for functions of one variable, it sounds like your explanation of changing from CCW to CW giving the maximum slope value, it seems that this applied not just to Trigonometric by all one-variable functions? Please let me know. $\endgroup$
    – Palu
    Mar 19, 2019 at 19:54
  • $\begingroup$ It would apply to any function that has a continuous 2nd derivative (on some interval.) $\endgroup$
    – B. Goddard
    Mar 19, 2019 at 19:57
  • $\begingroup$ Ok, so that is usually the case for Calculus taught at the Highschool level. So hence based on this, all one needs to do is set the second derivative of a function to zero, and that x-value gives , when plugged into the first derivative, the Maximum value of all the tangent slopes on the function. Let me know if i got this right. $\endgroup$
    – Palu
    Mar 19, 2019 at 20:03
  • $\begingroup$ Setting the 2nd derivative equal to zero and finding the solutions is the first step. But a particular point gives a max rate of change only if the rotation changes from CCW to CW. If it changes from CW to CCW, then it's a min. If it doesn't change, the it's not really an inflection point. $\endgroup$
    – B. Goddard
    Mar 19, 2019 at 20:07
  • $\begingroup$ OK, i realized the other rotation would give a minimum. But i am wondering if there is a direct algebraic way to determine if it changes from CCW to CW or from CW to CCW. I wonder if this is something that would be in the subject of Differential Geometry? $\endgroup$
    – Palu
    Mar 19, 2019 at 20:16
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The first derivative of any function $y=f(x)$ at point $x$ gives you the slope of the tangent at that point or the rate of change of the function $y=f(x)$ at the point $x$. If the derivative is positive or negative slope is positive or negative accordingly.

Now, the second derivative of the function $y=f(x)$ is the rate of change of the first derivative of the function $y=f(x)$, so if the second derivative is positive or negative then we can say the first derivative is increasing or decreasing continuously.

To get the maximum value of a function $y=f(x)$ observe that the slope of the tangent at that point is $0$ or the rate of change of the function itself is $0$. To get the maximum value of the rate of change $y'=f'(x)$, you need to have the slope of tangent at the point as $0$, for the function $y'=f'(x)$ which means $f''(x)=0$.

Hope this helps...

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