2 questions on parametric and implicit differntiations

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February 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
February 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 $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

b) set $\frac{dy}{dx}$ found in (a) equal to 0 and solve for t.
February 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 $y'$ then find $x'$ and to get $\tfrac{dy}{dx}$ form the fraction $\frac{y'}{x'}$ .

Quote:

b) Find the values of the parameter t at the points where dy/dx = 0

You want to solve $\frac{dy}{dx} = 0 \implies \frac{y'}{x'} = 0 \implies y' = 0$ .
Can you finish?
February 12th 2009, 12:38 PM
sharp357

ok i differentiated so dy/dx = cos 2t / - sin t

just wondering if this part is right?
February 12th 2009, 01:57 PM
skeeter

$y = \sin(2t)$

chain rule ...

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

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

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