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Parametric equations with trig fuctions

Hello, I was wondering if anyone could help me. I'm struggling with parametric equations with trig fuctions, but i think the root of it is that I cant seem to do trig equations anymore.

I have this questions which i have the wrong answer to so if anyone could tell me where im going wrong i would appriciate it.

1) A curve is defined by the parametric equations:

x = 3 + sin2t y = 8t + 4cos2t

a) Show that at the point Q, where t=0, the gradient of the curve is 4.

So firstly I did dx/dt = cos2t and then i did dy/dt = -4sin2t

So for dy/dx I got: -4sin2t/cos2t

So when i put t=0 into my calculator i got 0/0 = 0 hence i dont understand where im going wrong.

Can anyone help?

Thanks.
KaosLisa
Hello, I was wondering if anyone could help me. I'm struggling with parametric equations with trig fuctions, but i think the root of it is that I cant seem to do trig equations anymore.

I have this questions which i have the wrong answer to so if anyone could tell me where im going wrong i would appriciate it.

1) A curve is defined by the parametric equations:

x = 3 + sin2t y = 8t + 4cos2t

a) Show that at the point Q, where t=0, the gradient of the curve is 4.

So firstly I did dx/dt = cos2t and then i did dy/dt = -4sin2t

So for dy/dx I got: -4sin2t/cos2t

So when i put t=0 into my calculator i got 0/0 = 0 hence i dont understand where im going wrong.

Can anyone help?

Thanks.

Your differentiation is wrong.
dx/dt = 2cos2t
dy/dt = 8 - 8sin2t

dy/dx = dy/dt x 1/dx/dt
dy/dx = (8-8sin2t)/2cos2t

When t = 0
dy/dx = (8-8sin0)/2cos0 = (8-0)/2 = 4
Reply 2
Avatar for KaosLisa
KaosLisa
OP
Ah got it now thank you!

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