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AP Calculus AB : Implicit differentiation

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Example Question #31 : Implicit Differentiation

Find the normal line of the curve $\frac{x^{2}}{2}+y^{2}=4$ at the point $(1,\frac{\sqrt{14}}{2})$

Possible Answers:

$y=x\sqrt{14}+\frac{\sqrt{14}}{2}$

$y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

$y=x\sqrt{14}-{\sqrt{14}}$

$y=2x\sqrt{14}-\frac{\sqrt{14}}{2}$

Correct answer:

$y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

Explanation:

$\frac{x^{2}}{2}+y^{2}=4$

Use implicit differentiation to calculate the slope of the tangent line:

$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{x^{2}}{2}+y^{2}=4)$

$\frac{2x}{2}+2y\frac{dy}{dx}=0$

Simplify: $x+2y\frac{dy}{dx}=0$

Subtract x from both sides of the equation: $2y\frac{dy}{dx}=-x$

Divide both sides of the equation by 2y: $\frac{dy}{dx}=\frac{-x}{2y}$

Plug in the x and y values from the point $(1,\frac{\sqrt{14}}{2})$ into the equation, to calculate the slope of the tangent line: $\frac{dy}{dx}=\frac{-1}{\frac{2\sqrt{14}}{2}}=\frac{-1}{\sqrt{14}}=\frac{-\sqrt{14}}{14}$

Take the negative reciprocal of the slope of the tangent line to calculate the slope of the normal line: $m=-\frac{1}{\frac{-\sqrt{14}}{14}}=\sqrt{14}$

Plug the slope of the normal line into point slope form: $y-\frac{\sqrt{14}}{2}=(x-1)\sqrt{14}$

Add $\frac{\sqrt{14}}{2}$ to both sides to isolate y: $y=(x-1)\sqrt{14}+\frac{\sqrt{14}}{2}$

Simplify the equation: $y=x\sqrt{14}-\sqrt{14}+\frac{\sqrt{14}}{2}$

Simplify further: $y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

Solution: $y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

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AP Calculus AB : Implicit differentiation

Study concepts, example questions & explanations for AP Calculus AB

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Example Question #31 : Implicit Differentiation

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