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جستجو

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The parametric equations of a curve are

$x = ln\left( {\tan t} \right)$,  $y = {\sin ^2}{\text{ }}t$,

where $0 \lt t \lt \frac{1}{2}\pi $.

a) Express $\frac{{dy}}{{dx}}$ in terms of $t$.

b) Find the equation of the tangent to the curve at the point where $x = 0$.

پاسخ تشریحی :
نمایش پاسخ

a) EITHER: State $\frac{{dx}}{{dt}} = {\sec ^2}t/\tan t$, or equivalent

State $\frac{{dy}}{{dt}} = 2\sin t\cos t$, or equivalent

Use $\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \div \frac{{dx}}{{dt}}$

Obtain correct answer in any form, e.g. $2{\sin ^2}t{\cos ^2}t$

OR: Obtain $y = {e^{2x}}/\,\,\left( {1 + {e^{2x}}} \right)$, or equivalent

Use correct quotient or product rule

Obtain correct derivative in any form, e.g. $2{e^{2x}}/{\text{ }}{\left( {1 + {e^{2x}}} \right)^2}$

Obtain correct derivative in terms of $t$ in any form, e.g. $\left( {2{{\tan }^2}t} \right)/{\left( {1 + {{\tan }^2}t} \right)^2}$

b) State or imply $t = \frac{1}{4}\pi $ when $x = 0$

Form the equation of the tangent at $x = 0$

Obtain correct answer in any horizontal form, e.g. $y = \frac{1}{2}x + \frac{1}{2}$

[SR: If the $OR$ method is used in part (a), give B1 for stating or implying $y = \frac{1}{2}$ or $\frac{{dy}}{{dx}} = \frac{1}{2}$ when $x = 0$]

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