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find normal line equation of the curve y=sqrt(25-x^2) at the point (3,4)

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find normal line equation of the curve y=sqrt(25-x^2) at the point (3,4)

crysl  Mar 23, 2015

#2
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$$\\y=\sqrt{25-x^2}\\\\ y=(25-x^2)^{0.5}\\\\ \frac{dy}{dx}=0.5(25-x^2)^{-0.5}\times -2x\\\\ \frac{dy}{dx}=-x(25-x^2)^{-0.5}\\\\ when \;\;x=3\\\\ \frac{dy}{dx}=-3(25-3^2)^{-0.5}\\\\ \frac{dy}{dx}=-3(16)^{-0.5}\\\\ \frac{dy}{dx}=-3\times\frac{1}{4}\\\\ \frac{dy}{dx}=-\frac{3}{4}\\\\$$

$$\\Gradient of the tangent is  -\frac{3}{4}\\\\ Gradient of the normal is  +\frac{4}{3}\\\\\\ \frac{4}{3}=\frac{y-4}{x-3}\\\\ 4(x-3)=3(y-4)\\\\ 4x-12=3y-12\\\\ the equation of the normal at (3,4) is  4x-3y=0$$

Melody  Mar 23, 2015
#1
+86859
+10

y = (25 - x^2)^1/2    taking the derivative, we have

y' = (1/2)(25 -x^2)^(-1/2) (-2x) =  -x / √[25 - x^2]

And the slope of the line when x = 3 = -3/ √[25 - 3^2]  = -3/ √16 = -3/4

And the slope of the normal line will be the negative reciprocal of this = 4/3

So we have

y - 4 = (4/3)(x -3)

y = (4/3)x - 4 + 4

y = (4/3)x       and that's the equation of the normal line to the function at (3, 4)

See the pic here of the function, the tangent line to the function at (3, 4) and the normal line to the function going through the same point

https://www.desmos.com/calculator/ff3alev1sg

CPhill  Mar 23, 2015
#2
+92623
+10

$$\\y=\sqrt{25-x^2}\\\\ y=(25-x^2)^{0.5}\\\\ \frac{dy}{dx}=0.5(25-x^2)^{-0.5}\times -2x\\\\ \frac{dy}{dx}=-x(25-x^2)^{-0.5}\\\\ when \;\;x=3\\\\ \frac{dy}{dx}=-3(25-3^2)^{-0.5}\\\\ \frac{dy}{dx}=-3(16)^{-0.5}\\\\ \frac{dy}{dx}=-3\times\frac{1}{4}\\\\ \frac{dy}{dx}=-\frac{3}{4}\\\\$$

$$\\Gradient of the tangent is  -\frac{3}{4}\\\\ Gradient of the normal is  +\frac{4}{3}\\\\\\ \frac{4}{3}=\frac{y-4}{x-3}\\\\ 4(x-3)=3(y-4)\\\\ 4x-12=3y-12\\\\ the equation of the normal at (3,4) is  4x-3y=0$$

Melody  Mar 23, 2015