• October 3rd 2012, 04:23 PM
skinsdomination09
5) x = e^√t y= t - ln(t²) Correct Answer : 2√t(1-(2/t))/(e^√t)

8) x = 2sin2t y = 2 sint Correct Answer: cost/2cos2t *note I see the trig identitiy sin2t = 2cos*sin but I don't see how it helps

I get the concept that you have to put the derivitive of y over x to get the parametric derivitive but I just can't get the alrebra to work.
• October 3rd 2012, 05:17 PM
Soroban
Hello, skinsdomination09!

The algebra is not hard.

Quote:

$5)\;\begin{Bmatrix}x &=& e^{\sqrt{t}} \\ y&=& t -\ln(t^2)\end{Bmatrix}$

$\text{Answer: }\:\frac{2\sqrt{t}\left(1 - \frac{2}{t}\right)}{e^{\sqrt{t}}}$

$\text{We have: }\:\begin{Bmatrix}y &=& t - 2\ln(t) & \Rightarrow & \frac{dy}{dt} &=& 1 - \frac{2}{t} \\ x &=& e^{t^{\frac{1}{2}}} & \Rightarrow & \frac{dx}{dt} &=& e^{t^{\frac{1}{2}}}\!\cdot\!\frac{1}{2}t^{-\frac{1}{2}} \end{Bmatrix}$

$\frac{dy}{dx} \;=\;\frac{\frac{dy}{dt}}{\frac{dx}{dr}} \;=\; \frac{1-\frac{2}{t}}{\frac{e^{\sqrt{t}}}{2\sqrt{t}}} \;=\;\frac{2\sqrt{t}(1-\frac{2}{t})}{e^{\sqrt{t}}}$

Quote:

$8)\;\begin{Bmatrix}x &=& 2\sin2t \\ y &=& 2\sin t\end{Bmatrix}$

$\text{Answer: }\:\frac{\cos t}{2\cos2t}$

We have: . $\begin{Bmatrix}y &=& 2\sin t & \Rightarrow& \frac{dy}{dt} &=& 2\cos t \\ \\[-3mm] x &=& 2\sin2t & \Rightarrow & \frac{dx}{dt} &=& 4\cos2t \end{Bmatrix}$

Therefore: . $\frac{dy}{dx} \;=\;\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \;=\;\frac{2\cos t}{4\cos 2t} \;=\;\frac{\cos t}{2\cos2t}$