Find the derivative of the semicircle $y= \sqrt{169-x^2}$ and sketch the semi-circle

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We weren't taught this in class and I've tried to search it up but can't find the answer.

The derivative of $y=\sqrt{169 - x^2}$ is equal to $- x/\sqrt{169-x^2}$, but how do you find the radius of the semi-circle and sketch it?

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3 Answers

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Notice, we have $y=\sqrt{169-x^2}=(169-x^2)^{1/2}$ its derivative is determined by the chain rule as follows $$\frac{dy}{dx}=\frac{d}{dx}(169-x^2)^{1/2}=\frac{1}{2}(169-x^2)^{1/2-1}(-2x)=-x(169-x^2)^{-1/2}=-\frac{x}{\sqrt{169-x^2}}$$ Now, compare the given equation of circle : $$y=\sqrt{169-x^2}$$ $$y^2=169-x^2$$ $$x^2+y^2=169$$ with standard form: $x^2+y^2=r^2$ having center at the origin $(0, 0)$ & a radius $r$, thus we get $$\text{radius of semicircle }=\sqrt{169}=13$$

Since, you have not mentioned the interval of the variable $x$, hence we have

The semicircle will be on the right side of y-axis for all $0\leq x\leq 13$

The semicircle will be on the left side of y-axis for all $-13\leq x\leq 0$

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If $y = \sqrt{169-x^2}$, then $y^2 = 169 - x^2$. Can you rearrange this to get something that looks like Pythagoras theorem? That would tell you where $(x,y)$ must be to satisfy the equation.

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The equation of a semicircle is given by: $$y=\sqrt{r^2-x^2}$$ where $r$ is the radius. Hence in your case the radiu13*13 s is:$$\sqrt{169}=13$$

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