Tangent and Normal Lines at (1,0) on Curve y = pi*sin(pi*x-y)

[cummings15]

Homework Statement

Verify that (1,0) is on the following curve and find the tangent line and normal line to the curve at the point.

y=pisin(pix-y)

The Attempt at a Solution

i think i got it is y '
{-1/pi*cos(pi*x-y)} + pi

 

[Mark44]

Homework Statement

Verify that (1,0) is on the following curve and find the tangent line and normal line to the curve at the point.

y=pisin(pix-y)

This is difficult to read. Use * to indicate products, like so:
y = pi * sin(pi * x - y)

The Attempt at a Solution

y prime = picos(u)(du)

y prime = picos(pix-y)(pi-y prime)

Instead of writing y prime, you can use an apostrophe - ' - to indicate a derivative.

y ' = pi * cos(pi * x - y) * (pi - y')

This is correct as far as it goes. Use algebraic techniques to get all the terms involving y' on one side and everything else on the other. Solve for y' algebraically.

Did you show that (1, 0) is a point on the graph of this equation?

Once you have solved for y', find the equation of the tangent line by evaluating y' at (1, 0). That gives you the slope of the tangent line. For the equation of the tangent line, use the point-slope form of the equation of a line.

 

[cummings15]

updated first post

 

[cummings15]

updated first post is y ' right

 

[cummings15]

i think i got it is y '
{-1/pi*cos(pi*x-y)} + pi

 

[Dick]

i think i got it is y '
{-1/pi*cos(pi*x-y)} + pi

No, I don't think it is. Can you show us how you got that?

 

[cummings15]

i got this y ' = pi * cos(pi * x - y) * (pi - y')

and then i solved for y '

 

[Mark44]

i got this y ' = pi * cos(pi * x - y) * (pi - y')

and then i solved for y '

If you expand the right side, the equation becomes
y ' = pi² * cos(pi * x - y) - y' * pi * cos(pi * x - y)
Get all the terms involving y' on one side, and the rest on the other side, then solve for y'.

 

[cummings15]

so the slope would = pi

 

[Dick]

so the slope would = pi

No. Is that a guess? Why don't you substitute x=1 and y=0 into y ' = pi * cos(pi * x - y) * (pi - y') and solve for y'?