# Pls Chk Work (dx/dt)

• Aug 7th 2006, 04:52 PM
Yogi_Bear_79
Pls Chk Work (dx/dt)
Find dx/dt given x = sqrt(1 + cot 3t)?

dx/dt = 1/2 ( 1+cos3t) ^(-˝) [0 - cosec˛ 3t(3)]

dx/dt = 1/2{ 1/(1+cost3t)˝ } {-3cosec˛ 3t}

= [ (-3/2 ) (cosec˛3t) {1/(1+cos3t)˝} ]
• Aug 7th 2006, 05:08 PM
galactus
$\displaystyle \frac{-3csc^{2}(3t)}{2\sqrt{1+cot(3t)}}$

Looks good. :)
• Aug 7th 2006, 05:17 PM
Yogi_Bear_79
Could you convert the entire solution using the code you used? I am new to posting online and the example would be useful, and easier to read!
• Aug 7th 2006, 05:46 PM
galactus
Click quote and see what the code is. With a little practice you, too, can learn LaTex. I converted what you had.

Find $\displaystyle \frac{dx}{dt}$ given $\displaystyle x = \sqrt{1 + cot 3t}$?

$\displaystyle \frac{dx}{dt} = \frac{1}{2}( 1+cot(3t))^{\frac{-1}{2}}[0 - csc˛(3t)]$

$\displaystyle \frac{dx}{dt} = \frac{1}{2}\frac{1}{\sqrt{1+cot(3t)}}{(-3csc˛ (3t))}$
= $\displaystyle (\frac{-3}{2}) (csc˛(3t)) {\frac{1}{\sqrt{1+cot(3t)}}$

=$\displaystyle \frac{-3csc^{2}(3t)}{2\sqrt{1+cot(3t)}}$