A curve is given parametrically by the equation x = 3-4sinT and y = 4+3cosT for 0 <= T <= 2pi. What are all the points (x,y) at which the curve has a vertical tangent?

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- Apr 7th 2008, 09:19 AMDINOCALC09parametric: points of curve that has vert tangent
A curve is given parametrically by the equation x = 3-4sinT and y = 4+3cosT for 0 <= T <= 2pi. What are all the points (x,y) at which the curve has a vertical tangent?

- Apr 7th 2008, 09:21 AMMoo
Hello,

Calculate dy/dx, then find t as dy/dx tends to infinity - Apr 7th 2008, 09:26 AMDINOCALC09
so d/dx = -4cosT

and

d/dy = -3sinT

then wat - Apr 7th 2008, 09:34 AMMoo
dx/dt=-4cost

dy/dt=-3sint

so dy/dx=dy/dt * dt/dx

dy/dx=-3sin(t)/(-4cos(t))

So it's infinite when cos(t)=0, with t between 0 and 2pi

cos(pi/2)=cos(3pi/2)=0

So at t=pi/2 and t=3pi/2, there is a vertical tangent.

t=pi/2 -> x = 3-4sin (pi/2)=3-4=-1, y = 4+3cos(pi/2)=4 ------> (-1,4)

t=3pi/2 -> x = 3-4sin (3pi/2)=3-4*(-1)=7, y = 4+3cos(3pi/2) = 4 ------> (7,4)