1.

Find the equation of the tangent and normal to the curve y= 1 + sin 2x at the point x=$\displaystyle \frac{\pi}{12}$.

2.

The volume of a cone V $\displaystyle cm^3$ is given by the formula V= $\displaystyle kr^3 cot \theta$ , where k is constant, r is the radius of the cone and $\displaystyle \theta$ is a semi vertical angle. Given that r is fixed and $\displaystyle \theta$ is increasing at a rate of 0.2 radian per second, calculate the rate of change of volume with respect to time when $\displaystyle \theta$ = $\displaystyle \frac{\pi}{6}$.

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i think i know how to do, but the problem ismy answer is not the same as the book's answer. Can someone please tell me where i went wrong ??

1.

$\displaystyle \frac{dy}{dx} = -2 cos 2x$

When x= $\displaystyle \frac{\pi}{12}$, $\displaystyle \frac{dy}{dx} = -\sqrt3$

When x= $\displaystyle \frac{\pi}{12}$, y= $\displaystyle \frac{3}{2}$

equation of tangent :

$\displaystyle y - \frac{3}{2} = - \sqrt3 (x - \frac{\pi}{12} )$

$\displaystyle 2y -3 = -2 \sqrt3 x + \frac{2 \sqrt3 \pi}{12}$

$\displaystyle 2y= -2 \sqrt3 x + 3 + \frac{\sqrt3 \pi}{6}$

the book's answer is $\displaystyle 2y= 2 \sqrt3 x +3 - \frac{\sqrt3 \pi}{6}$

so my answer is different from the book's answer in terms of the positive sign and negative sign .....

gradient of normal= $\displaystyle \frac{1}{\sqrt3}$

equation of normal:

$\displaystyle y- \frac{3}{2} = \frac{1}{\sqrt3} (x - \frac{\pi}{12})$

$\displaystyle 2y-3 = \frac{2}{\sqrt3} x - \frac{2\pi}{12 \sqrt3}$

$\displaystyle 2y= \frac{2}{\sqrt3} x + 3 - \frac{\pi}{6 \sqrt3}$

$\displaystyle 2 \sqrt3 y - 2x = 3 \sqrt3 - \frac{\pi}{6}$

the book's answer is $\displaystyle 2 \sqrt3 y + 2x = 3 \sqrt3 + \frac{\pi}{6}$

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2.

V = $\displaystyle kr^3 cot \theta$

= $\displaystyle kr^3 (tan \theta)^{-1}$

$\displaystyle \frac{dv}{d \theta} $

=$\displaystyle - kr^3 (tan \theta)^{-2} sec^2 \theta$

=$\displaystyle - kr^3 sec^4 \theta$

$\displaystyle \frac{d \theta}{dt} = 0.2$

Find $\displaystyle \frac{dV}{dt}$ when $\displaystyle \theta = \frac{\pi}{6}$ .

$\displaystyle \frac{dV}{dt} = \frac{d \theta}{dt} * \frac{dV}{d \theta}$

= 0.2 * ( $\displaystyle - kr^3 sec^4 \frac{\pi}{6}$)

=$\displaystyle ( - \frac{kr^3}{5})(\frac{1}{tan \frac{\pi}{6}}^4$

=$\displaystyle ( - \frac{kr^3}{5})(\frac{3}{\sqrt3} )^4$

=( - $\displaystyle \frac{kr^3}{5}$)($\displaystyle \frac{81}{9}$)

=$\displaystyle - \frac{9kr^3}{5} $

the book's answer is $\displaystyle - \frac{4kr^3}{5}$.....

Can someone please tell me where i went wrong ?? Thanks.