find an equation for the line tangent to the curve

[sareen]

for the parametric curve defined by x=2sin(t) and y=2cos(t)
a) Find the length of the curve for 0?t? ?/2
answer) L=?

Now for part b) find an equation for the line tangent to the curve at the point where t=? /4

How to solve this???

 

[galactus]

for the parametric curve defined by x=2sin(t) and y=2cos(t)
a) Find the length of the curve for 0?t? ?/2
answer) L=?

Just use the arc length formula.

$\displaystyle \int_{0}^{\frac{\pi}{2}}\sqrt{\left(\frac{d}{dt}(2sin(t))\right)^{2}+\left(\frac{d}{dt}(2cos(t))\right)^{2}}dt$

This whittles down to a very easy integral.

Now for part b) find an equation for the line tangent to the curve at the point where t=? /4

How to solve this???

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

Then, plug in $\displaystyle t=\frac{\pi}{4}$ to find the slope at that point. You have $\displaystyle y=2cos(\frac{\pi}{4}), \;\ x=2sin(\frac{\pi}{4})$Then, use y=mx+b and solve for b.

 

[BigGlenntheHeavy]

$\displaystyle Another \ way, \ reverting \ to \ rectangular \ coordinates.$

$\displaystyle x \ = \ 2sin(t), \ y \ =2cos(t) \ \implies \ x^{2}+y^{2} \ = \ 4$

$\displaystyle a) \ C \ = \ 2\pi r \ = \ 4\pi. \ \frac{C}{4} \ = \ \frac{4\pi}{4} \ = \ \pi.$

$\displaystyle b) \ At \ \pi/4 \ m \ =1 \ and \ we \ have \ the \ point \ (\sqrt2,\sqrt2)$

$\displaystyle The \ tangent \ line \ is \ perpendicular \ to \ the \ the \ line \ emanating \ from \ the \ origin, \ so \ m \ = \ -1$

$\displaystyle with \ point \ (\sqrt2,\sqrt2) \ gives \ equation \ y \ = \ -x+2\sqrt2, \ see \ graph.$

[attachment=0:37tv7sqs]ghi.jpg[/attachment:37tv7sqs]