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For graphing the derivative of the circle, I know that the equation of a circle is $x^2+y^2 = r^2$ and in this case r = 4

With implicit differentiation I know that $y' = \frac{-x}{y}$ or $\frac{-x}{\sqrt{16-x^2}}$

I need to graph this derivative from $-6 \leq x \leq -2$

When I plug it into a graphing calculator (DESMOS) I get the graph as this:

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More importantly, from -6 to -2, I have this :

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Can someone explain how I can graph this derivative.

Also, this is 1/4 of the circle, how does it make a difference in the question (like what part of my work does it affect?)

  • $\begingroup$ What do you mean by graphing the derivative? Judging from the pictures, you have already graphed it. $\endgroup$ – Servaes Jan 21 '19 at 13:59
  • $\begingroup$ Also, judging from the question in your first picture, the equation of your circle segment is not $x^2+y^2=4$ because the circle is not centered at the origin. $\endgroup$ – Servaes Jan 21 '19 at 14:02

Judging from the question in the first picture, it seems that you were asked to sketch the derivative of the quarter circle. The drawn correction suggests that it suffices to note that the derivative is $0$ at the left of the arc, and decreases monotonically to negative infinity towards the right of the arc.

This can be made more precise; the arc is given by the equation $$(x+6)^2+y^2=16,$$ so implicit differentiation yields the derivative $$y'=-\frac{x+6}{\sqrt{16-(x+6)^2}}.$$ Plugging in $x=-6$ indeed yields $y'=0$, and it is not hard to check that $\lim_{x\to-2}y'=-\infty$. To verify that $y'$ is monotonically decreasing you can verify that the second derivative $$y''=-\frac{16}{\sqrt{16-(x+6)^2}^3},$$ is negative on the interval $(-6,-2)$. Moreover, plugging in a few values for $x$ into the expression for $y'$ allows you to draw the curve more precisely. For example, $$y'(4)=-\frac{10}{\sqrt{16-2^2}}=-\frac{5}{\sqrt{3}}\approx-2.88675...$$ Though judging from the drawn correction, such precision is not necessary.

  • $\begingroup$ how do you know that the derivative decreases monotonically to negative infinity towards the right of the arc? and also, what is wrong with my solution? $\endgroup$ – user8290579 Jan 21 '19 at 14:13
  • $\begingroup$ You can 'see' it decreases monotonically because the slope of the curve decreases more and more rapidly as you move to the right. This can be made rigorous by computing the second derivative. I have added some more details, which complement your approach but don't seem to be necessary to answer the question. $\endgroup$ – Servaes Jan 21 '19 at 14:15
  • $\begingroup$ One part of your solution that is wrong is working with the equation $x^2+y^2=r^2$. This is an equation for a circle centered at the origin. $\endgroup$ – Servaes Jan 21 '19 at 14:18
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    $\begingroup$ thank you! just one question about the second derivative, you are saying it is positive on (-6,-2), wouldn't that make the function concave up? but the function is concave down on that interval $\endgroup$ – user8290579 Jan 21 '19 at 14:26
  • $\begingroup$ Oh I overlooked the minus-sign, it is negative on the whole interval. I'll edit that right now. $\endgroup$ – Servaes Jan 21 '19 at 14:33

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