Finding an equation tangent line

Hello everyone

How would i solve the following

Find an equation for the tangent line to the graph of x+sin(y-2x)=1 at the point (1,2)

I did

1+cos(y-2x)(dy/dx)(-2)=0

then I did cos(y-2x)(dy/dx)=2

(dy)/(dx)=(2)/(cos(y-2x) so would I just plug in the value of x and y.

Re: Finding an equation tangent line

That would give you the slope. You need the equation for the tangent line. The equation of a line through $\displaystyle (x_0,y_0)$ with slope m is $\displaystyle (y-y_0)=m(x-x_0)$.

- Hollywood

Re: Finding an equation tangent line

Quote:

Originally Posted by

**homeylova223** Hello everyone

How would i solve the following

Find an equation for the tangent line to the graph of x+sin(y-2x)=1 at the point (1,2)

I did

1+cos(y-2x)(dy/dx)(-2)=0

then I did cos(y-2x)(dy/dx)=2

This is incorrect. from 1+ cos(y- 2x)(dy/dx)(-2)= 0 you get -2cos(y- 2x)(dy/dx)= -1 or cos(y- 2x)(dy/dx)= 1/2. When x= 1, y= 2, cos(2- 1)(dy/dx)= 1/2 so dy/dx= 1/cos(1).

Quote:

(dy)/(dx)=(2)/(cos(y-2x) so would I just plug in the value of x and y.

Re: Finding an equation tangent line

the Gradient is 1 and the equation of the tangent is x-y=-1.

Re: Finding an equation tangent line

There is a slight problem, when you differentiate you will get

1 + cos ( y - 2x ) [ dy/dx -2] = 0

At ( 1,2 ) we get dy/dx = 1

So the equation of tangent is ( y-2) = 1 * ( x-1)

OR

x - y + 1 = 0

Re: Finding an equation tangent line

Yeah, $\displaystyle \frac{dy}{dx}=1$. Sorry I missed that.

- Hollywood