1. ## questions about differentiation of trigo

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

2.
The volume of a cone V $cm^3$ is given by the formula V= $kr^3 cot \theta$ , where k is constant, r is the radius of the cone and $\theta$ is a semi vertical angle. Given that r is fixed and $\theta$ is increasing at a rate of 0.2 radian per second, calculate the rate of change of volume with respect to time when $\theta$ = $\frac{\pi}{6}$.

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

1.
$\frac{dy}{dx} = -2 cos 2x$
When x= $\frac{\pi}{12}$, $\frac{dy}{dx} = -\sqrt3$
When x= $\frac{\pi}{12}$, y= $\frac{3}{2}$

equation of tangent :
$y - \frac{3}{2} = - \sqrt3 (x - \frac{\pi}{12} )$
$2y -3 = -2 \sqrt3 x + \frac{2 \sqrt3 \pi}{12}$
$2y= -2 \sqrt3 x + 3 + \frac{\sqrt3 \pi}{6}$

the book's answer is $2y= 2 \sqrt3 x +3 - \frac{\sqrt3 \pi}{6}$
so my answer is different from the book's answer in terms of the positive sign and negative sign .....

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

equation of normal:
$y- \frac{3}{2} = \frac{1}{\sqrt3} (x - \frac{\pi}{12})$
$2y-3 = \frac{2}{\sqrt3} x - \frac{2\pi}{12 \sqrt3}$
$2y= \frac{2}{\sqrt3} x + 3 - \frac{\pi}{6 \sqrt3}$
$2 \sqrt3 y - 2x = 3 \sqrt3 - \frac{\pi}{6}$

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

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2.
V = $kr^3 cot \theta$
= $kr^3 (tan \theta)^{-1}$

$\frac{dv}{d \theta}$
= $- kr^3 (tan \theta)^{-2} sec^2 \theta$
= $- kr^3 sec^4 \theta$

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

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

$\frac{dV}{dt} = \frac{d \theta}{dt} * \frac{dV}{d \theta}$
= 0.2 * ( $- kr^3 sec^4 \frac{\pi}{6}$)
= $( - \frac{kr^3}{5})(\frac{1}{tan \frac{\pi}{6}}^4$
= $( - \frac{kr^3}{5})(\frac{3}{\sqrt3} )^4$
=( - $\frac{kr^3}{5}$)( $\frac{81}{9}$)
= $- \frac{9kr^3}{5}$

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

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

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

2.
The volume of a cone V $cm^3$ is given by the formula V= $kr^3 cot \theta$ , where k is constant, r is the radius of the cone and $\theta$ is a semi vertical angle. Given that r is fixed and $\theta$ is increasing at a rate of 0.2 radian per second, calculate the rate of change of volume with respect to time when $\theta$ = $\frac{\pi}{6}$.

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

1.
$\frac{dy}{dx} = -2 cos 2x$
When x= $\frac{\pi}{12}$, $\frac{dy}{dx} = -\sqrt3$
When x= $\frac{\pi}{12}$, y= $\frac{3}{2}$

equation of tangent :
$y - \frac{3}{2} = - \sqrt3 (x - \frac{\pi}{12} )$
$2y -3 = -2 \sqrt3 x + \frac{2 \sqrt3 \pi}{12}$
$2y= -2 \sqrt3 x + 3 + \frac{\sqrt3 \pi}{6}$

the book's answer is $2y= 2 \sqrt3 x +3 - \frac{\sqrt3 \pi}{6}$
so my answer is different from the book's answer in terms of the positive sign and negative sign .....

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

equation of normal:
$y- \frac{3}{2} = \frac{1}{\sqrt3} (x - \frac{\pi}{12})$
$2y-3 = \frac{2}{\sqrt3} x - \frac{2\pi}{12 \sqrt3}$
$2y= \frac{2}{\sqrt3} x + 3 - \frac{\pi}{6 \sqrt3}$
$2 \sqrt3 y - 2x = 3 \sqrt3 - \frac{\pi}{6}$

the book's answer is $2 \sqrt3 y + 2x = 3 \sqrt3 + \frac{\pi}{6}$
$y= 1 + sin 2x$
$\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \cos 2x$

If you are differentiating $\cos x$ then you get $-\sin x$ (Negative) but if you are differentiating $\sin x$ then you get $\cos x$ (Positive).

Originally Posted by wintersoltice
======================

2.
V = $kr^3 cot \theta$
= $kr^3 (tan \theta)^-1$

$\frac{dv}{d \theta}$
= $- kr^3 (tan \theta)^(-2) sec^2 \theta$
= $- kr^3 sec^4 \theta$

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

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

$\frac{dV}{dt} = \frac{d \theta}{dt} * \frac{dV}{d \theta}$
= 0.2 * ( $- kr^3 sec^4 \frac{\pi}{6}$)
= $( - \frac{kr^3}{5})(\frac{1}{tan \frac{\pi}{6}}^4$
= $( - \frac{kr^3}{5})(\frac{3}{\sqrt3} )^4$
=( - $\frac{kr^3}{5}$)( $\frac{81}{9}$)
= $- \frac{9kr^3}{5}$

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

Can someone please tell me where i went wrong ?? Thanks.
You correctly wrote that $\cot x = \frac{1}{\tan x}$ but further down you changed your $\frac{1}{\tan ^2 x}$ to $\sec ^2 x$ which is incorrect.

$\frac{\mathrm{d}v}{\mathrm{d} \theta} = - kr^3 (\tan ^{-2} \theta)\sec ^2 \theta = -\frac{kr^3}{\tan ^2\theta \cos ^2 \theta}$