# 2 questions on parametric and implicit differntiations

• Feb 11th 2009, 12:39 PM
sharp357
2 questions on parametric and implicit differntiations
Ok the curve has parametric equations:

x = cos t y = sin 2t, 0 ≤ t < 2pi

a) Find an expression for dy/dx in terms of parameter t
b) Find the values of the parameter t at the points where dy/dx = 0

An example in my book has confused me and so i was wondering if somebody could show me how to work through these two questions.
Thanks
• Feb 11th 2009, 02:23 PM
skeeter
Quote:

Originally Posted by sharp357
the curve has parametric equations:

x = cos t y = sin 2t, 0 ≤ t < 2pi

a) Find an expression for dy/dx in terms of parameter t
b) Find the values of the parameter t at the points where dy/dx = 0

thanks

a) note that $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

b) set $\displaystyle \frac{dy}{dx}$ found in (a) equal to 0 and solve for t.
• Feb 11th 2009, 02:58 PM
ThePerfectHacker
Quote:

Originally Posted by sharp357
Ok the curve has parametric equations:

x = cos t y = sin 2t, 0 ≤ t < 2pi

a) Find an expression for dy/dx in terms of parameter t

First find $\displaystyle y'$ then find $\displaystyle x'$ and to get $\displaystyle \tfrac{dy}{dx}$ form the fraction $\displaystyle \frac{y'}{x'}$.

Quote:

b) Find the values of the parameter t at the points where dy/dx = 0
You want to solve $\displaystyle \frac{dy}{dx} = 0 \implies \frac{y'}{x'} = 0 \implies y' = 0$.
Can you finish?
• Feb 12th 2009, 12:38 PM
sharp357
ok i differentiated so dy/dx = cos 2t / - sin t

just wondering if this part is right?
• Feb 12th 2009, 01:57 PM
skeeter
$\displaystyle y = \sin(2t)$

chain rule ...

$\displaystyle \frac{dy}{dt} = 2\cos(2t)$

$\displaystyle \frac{dx}{dt} = -\sin{t}$ is correct