# A curve has parametric equations x= 2t-ln 2t and y= t² - ln t² where t>0 . Find the value of t at the point on the curve where dy/dx=2 . Hence find the coordinates of that point?

Aug 28, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2$ when $t = 2$. The coordinates of the corresponding point are $\left(x , y\right) = \left(x \left(2\right) , y \left(2\right)\right) = \left(4 - \ln \left(4\right) , 4 - \ln \left(4\right)\right) \approx \left(2.61 , 2.61\right)$.

#### Explanation:

We have $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 - \frac{1}{t} = \frac{2 t - 1}{t}$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - \frac{2}{t} = \frac{2 {t}^{2} - 2}{t}$. Therefore $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{2 {t}^{2} - 2}{2 t - 1} = \frac{2 \left({t}^{2} - 1\right)}{2 t - 1}$.

Setting $\frac{\mathrm{dy}}{\mathrm{dx}} = 2$ results in $\frac{{t}^{2} - 1}{2 t - 1} = 1$, or ${t}^{2} - 1 = 2 t - 1$, or ${t}^{2} - 2 t = t \left(t - 2\right) = 0$. The solutions of this equation are $t = 0 , 2$. However, $t = 0$ is not part of the domain ($t > 0$) of the original parametric curve.

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = 2$ when $t = 2$.

Since $x \left(2\right) = 4 - \ln \left(2\right)$ and $y \left(2\right) = 4 - \ln \left(2\right)$, the coordinates of the corresponding point are $\left(x , y\right) = \left(x \left(2\right) , y \left(2\right)\right) = \left(4 - \ln \left(4\right) , 4 - \ln \left(4\right)\right) \approx \left(2.61 , 2.61\right)$.

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