C3 differentiation

Hey, am I mean to get an answer with π in for this....:

The tangent to the curve y=tan2x at the point where x =π/8 meets the y-axis at the point Y. Find the distance oy where O is the origin.

So I differentiated it and found the equation of the tangent , but using the cartesian format, c = something with a π in it ... is this right. I would type out my working but I threw away the paper it was on by mistake - oh what is the world coming to, I shall have to work through it again, poo. Anyways, so is it meant to have a π in it or not ??!!

franks xxx

Reply 1

Cexy

Right.

y = tan2x

So dy/dx = 2sec²2x

When x = π/8, we have dy/dx = 4 (check this) and y = 1 (check this too).

So the tangent we want has the equation y = 4x + c, and we know that it goes through the point (π/8, 1). Subbing this in gives:

1 = 4π/8 + c, therefore c = 1 - π/2.

This meets the x-axis when y = 0, i.e. when 0 = 4x +1 - π/2, and so the point Y is x = π/8 - 1/4, y = 0.

It's obvious that thew distance OY is then π/8 - 1/4.

DISCLAIMER: I strongly suggest that you work through this yourself as I am very prone to making mistakes, especially when I'm typing stuff like this out from my head!

Reply 2

Twiglet

This meets the x-axis when y = 0, i.e. when 0 = 4x +1 - π/2, and so the point Y is x = π/8 - 1/4, y = 0.

It's obvious that thew distance OY is then π/8 - 1/4.

but Y is where it meets the y axis
I agree with your working up to this point, (I have found the paper I wrote my answer on - it was an envelope thats why I couldnt find it before!) But then I have done :

when x = pi/8 y=1
so 1= 4pi/8 + c
so c = (2-pi)/2 ...... this is the same as c = 1 - π/2. which is what you got half way through.... I am rying to work out the right thing arent I ?!

Reply 3

steve2005

franks

Hey, am I mean to get an answer with ? in for this....:

The tangent to the curve y=tan2x at the point where x =?/8 meets the y-axis at the point Y. Find the distance oy where O is the origin.

Here is a graph... I don't know how far you have got?