college lever calculus

Question

Find an equation of the tangent line to the curve

cos(x−y) =e ^-XY-1 at the point P (π / 2,0).

Craft your response in the form of a letter to a friend, giving brief explanations for each part of the computation.

Answer 1

Dear friend,

To find the line tangent to cos(x-y) = e^-xy-1, we need to get a line equation in the slope-intercept form for the point, P(π/2,0).

The slope, m, will be dy/dx or y'. However, we need to use implicit derivation to get y' from the curve equation.

So, doing implicit derivation, using chain and product rules,

-sin(x-y)(1-y') = e^-xy(-y - xy')

After some manipulation:

sin(x-y)y' - sin(x-y) = -ye^-xy - xe^-xyy'

Putting the y' terms on the left hand side of the equal sign and the rest on the right hand side, gives:

(sin(x-y) + xe^-xy) y' = sin(x-y) - ye^-xy

and solving for y':

y' = (sin(x-y) -ye^-xy)/(sin(x-y)+xe^-xy)

To get the slope, m, substitute in x = π/2 and y = 0:

m = sin(π/2)/(sin(π/2) + π/2) = 1/(1+π/2) = 2/(2+π)

Now we need to finish the equation for the tangent line:

y-0 = m(x-π/2) = 2x/(2+π) - π/(2+π)

y = 2x/(2+π) - π/(2+π)

Sincerely,

Your friend