What is the equation of the tangent line of #r=tan^2(theta) - sin(theta-pi)# at #theta=pi/4#?

Question

Luke Alvarez · Accepted Answer

#r=(2+sqrt2)/2# #r=tan^2 theta- sin(theta - pi)# at #pi/4# #r=tan^2 (pi/4) - sin (pi/4 -pi)# #r=1^2 - sin((-3pi)/4)# #r=1-sin((5pi)/4)# #r=1-(-sqrt2/2)# #r=1+sqrt2/2# #r=(2+sqrt2)/2#

Samantha Adkison · Answer

undefined To find the equation of the tangent line, we first need to find the derivative of the given function with respect to θ, and then evaluate it at θ = π/4.

The given function is r = tan^2(θ) - sin(θ - π).

Differentiating with respect to θ, we get:

dr/dθ = d/dθ (tan^2(θ) - sin(θ - π))
      = 2tan(θ)sec^2(θ) - cos(θ - π)

Now, evaluate dr/dθ at θ = π/4:

dr/dθ = 2tan(π/4)sec^2(π/4) - cos(π/4 - π)
      = 2(1)(2^2) - (-√2/2)
      = 4 - (-√2/2)
      = 4 + √2/2
      = (8 + √2)/2

Now, we need to find the value of r at θ = π/4:

r(π/4) = tan^2(π/4) - sin(π/4 - π)
        = (1)^2 - sin(π/4 - π)
        = 1 - sin(π/4 - π)
        = 1 - sin(-3π/4)
        = 1 - (-√2/2)
        = 1 + √2/2

So, the coordinates of the point where the tangent line touches the curve are (π/4, 1 + √2/2).

Now, we can write the equation of the tangent line using the point-slope form:

y - y1 = m(x - x1)

where (x1, y1) = (π/4, 1 + √2/2) and m = (8 + √2)/2.

Thus, the equation of the tangent line is:

r - (1 + √2/2) = ((8 + √2)/2)(θ - π/4)