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?

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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?

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Answer 1 · 2015-08-28T01:47:53

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

Explanation:

We have $\frac{d x}{d t} = 2 - \frac{1}{t} = \frac{2 t - 1}{t}$ $dx/dt=2-1/t=(2t-1)/t$ and $\frac{d y}{d t} = 2 t - \frac{2}{t} = \frac{2 t^{2} - 2}{t}$ $dy/dt=2t-2/t=(2t^2-2)/t$ . Therefore $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 t^{2} - 2}{2 t - 1} = \frac{2 (t^{2} - 1)}{2 t - 1}$ $dy/dx=(dy/dt)/(dx/dt)=(2t^2-2)/(2t-1)=(2(t^2-1))/(2t-1)$ .

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

Therefore, $\frac{d y}{d x} = 2$ $dy/dx=2$ when $t = 2$ $t=2$ .

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

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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?

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