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Find the slope of the normal to the curv...

Find the slope of the normal to the curve `x = a cos^(3)theta, y=a sin^(3)theta` at `theta = (pi)/(4)`.

Text Solution

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`x = acos^3theta, y = asin^3theta`
`=>dx/(d theta) = -3acos^2thetasintheta`
`dy/(d theta) = 3asin^2thetacostheta`
`:. dy/dx = (dy/(d theta))/(dx/(d theta)) = (3asin^2thetacostheta)/(-3acos^2thetasintheta) = -tantheta`
`dy/dx|_(theta = pi/4) = -tanpi/4 = -1`
So, slope of tangent at `theta = pi/4` is `-1`.
Let slope of normal is `m`.
Then, `m*(-1) = -1`
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