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  • Find the slope of the normal to the curve x is equal to a cos raised to 3 theta , y = a sin raised to 3 theta at theta is equal to pi by 4 .

5) Find the slope of the normal to the curve x = a \cos ^3 \theta , y = a\sin ^3 \theta \: \: at \: \: \theta = \pi /4

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The slope of the tangent at a point on a given curve is given by 
 \left ( \frac{dy}{dx} \right )
Now,
\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\cos^2 \theta(-\sin \theta) = 3a(\frac{1}{\sqrt2})^2(-\frac{1}{\sqrt2}) = -\frac{3\sqrt2 a}{4}
Similarly,
\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\sin^2 \theta(\cos \theta) = 3a(\frac{1}{\sqrt2})^2(\frac{1}{\sqrt2}) = \frac{3\sqrt2 a}{4}
\left ( \frac{dy}{dx} \right ) = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{\frac{3\sqrt2 a}{4}}{-\frac{3\sqrt2 a}{4}} = -1
Hence, the slope of the tangent at  \theta = \frac{\pi}{4}      is -1
Now,
Slope of normal =  -\frac{1}{slope \ of \ tangent} = -\frac{1}{-1} = 1
Hence, the slope of normal at \theta = \frac{\pi}{4}       is 1

 

Posted by

Gautam harsolia

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