urgent help....

**tiga killa** · March 23rd 2009, 01:28 PM

Write the equation of the tangent line to the parametric curve x=sint y=sect when t=pi/4

SET UP the integral to calculate the arc length of the curve if t∈[0,pi/3]

**Pinkk** · March 23rd 2009, 01:44 PM

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{sec(t)tan(t)}{cos(t)}=\frac{sin(t)}{cos^{3}( t)}$

$\frac{dy}{dx}(\frac{\pi}{4})=\frac{\frac{\sqrt{2}} {2}}{(\frac{\sqrt{2}}{2})^{3}}=(\frac{\sqrt{2}}{2} )^{-2}=2$

^That is the slope of your tangent.

$x(\frac{\pi}{4})=\frac{\sqrt{2}}{2},y(\frac{\pi}{4 })=\sqrt{2}$

The equation of your tangent line in the form $y-b=m(x-a)$ :

$y-\sqrt{2}=2(x-\frac{\sqrt{2}}{2})$

Arc length:

$\int_0^{\frac{\pi}{3}}\sqrt{[x'(t)]^{2}+[y'(t)]^{2}}\,\,\,dt$

$\int_0^{\frac{\pi}{3}}\sqrt{cos^{2}(t)+sec^{2}(t)t an^{2}(t)}\,\,\,dt$

**Soroban** · March 23rd 2009, 02:01 PM

Hello, tiga killa!

Write the equation of the tangent line to the parametric curve
. . $x\:=\:\sin t,\;\;y\:=\:\sec t \:\text{ when }t\,=\,\tfrac{\pi}{4}$

When $t = \tfrac{\pi}{4} \!:\;\begin{Bmatrix}x \:=\:\sin\frac{\pi}{4} \:=\:\frac{\sqrt{2}}{2} \\ y \:=\:\sec\frac{\pi}{4} \:=\:\sqrt{2} \end{Bmatrix}\quad \Rightarrow \text{ The point is: }\:\left(\tfrac{\sqrt{2}}{2},\:\sqrt{2}\right)$

The derivatives are: . $\begin{Bmatrix}\dfrac{dy}{dt} \:=\:\sec t\tan t \\ \dfrac{dx}{dt} \:=\:\cos t \end{Bmatrix} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:\frac{\sec t\tan t}{\cos t} \:=\:\sec^2\!t\tan t$

When $t = \tfrac{\pi}{4}\!:\;\;\frac{dy}{dx} \;=\;\sec^2\tfrac{\pi}{4}\tan\tfrac{\pi}{4} \;=\;(\sqrt{2})^2(1) \;=\;2$

Now write the equation of the line through $\left(\tfrac{\sqrt{2}}{2},2\right)$ with slope 2.

SET UP the integral to calculate the arc length of the curve if $t \in \left[0,\,\tfrac{\pi}{3}\right]$

Formula: . $L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt$

Therefore: . $L \;=\;\int^{\frac{\pi}{3}}_0\sqrt{\cos^2\!t + \sec^2\!t\tan^2\!t}\:dt$

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