Calculus answer check.?

Question

the question asks, "Find all points (r, (pheta)) at which there is a vertical tangent line to the graph of r=1+sin (pheta)." Is the answer "x=cos(pheta) + sin (pheta)" ?

Answer 1

You mean "theta"? They're asking for all the points (r, θ) where this happens, so I don't see how an expression for x like that can be an answer.

f(θ) = r = 1 + sin(θ)
f ' (θ) = cos(θ)

dy/dx = (dy/dθ) / (dx/dθ)
This is vertical when (dx/dθ) = 0

dx/dθ = f '(θ) cos(θ) - f(θ) sin(θ)
dx/dθ = cos^2(θ) - (1+sin(θ))sin(θ)
dx/dθ = cos^2(θ) - sin(θ) - sin^2(θ)
dx/dθ = 1 - sin^2(θ) - sin(θ) - sin^2(θ)

-2sin^2(θ) - sin(θ) + 1 = 0
2sin^2(θ) + sin(θ) - 1 = 0
Let p = sin(θ)
2p^2 + p - 1 = 0
p = [ -1±√(1 - 4(2)(-1) ] / 4
p = [ -1±√9 ] / 4
p = -4/4, 2/4
so sin(θ) = -1, 1/2
This means θ = 270, 30, 150
and r = 1+-1, 1+1/2, 1+1/2
So the points are (0, 270), (3/2, 30), (3/2, 150).

Answer 2

This curve is given in polar coodinates. We have

y = r sin(pheta), x = r cos(pheta), so

y = sin(pheta) + sin^2(pheta) and x = cos(pheta) + sin(pheta) cos(pheta) = cos(pheta) + 1/2 sin(2 pheta).

It follows that

dy/dpheta = cos(pheta) + sin(2pheta) and
dx/dpheta = -sin(pheta) + cos(2 pheta)

The tangent line to the graph is vertical if dx/dpheta = 0, tha is , if

-sin(pheta) + cos(2 pheta) = 0 => cos(2 pheta) = sin(pheta). In [0, 2pi), this happens if and only if

2 pheta = pi/2 - pheta => pheta = pi/6
or
pheta = 2 pheta + pi/2 => pheta = -pi/2

For pheta = pi/6. dy/dpheta <>0, so that we actually have a vertical tangent line. But for pheta = -pi/2, dy/dpheta =0

At pheta = -pi/2 the slope of the tangent line is dy/dx = lim (pheta --> -pi/2) (cos(pheta) + sin(2pheta))/(sin(pheta) + cos(2 pheta)) = lim (x --> -pi/2) (-sin(pheta) + 2cos(2pheta)/(cos(pheta - 2sin(2pheta), using L'Hopital. When pheta --> -pi/2, the denominator --> -1 -2 = -3 and the denominator --> 0 - 2* 0 = 0. So, we, again, have a vertical tangnet line

The answer is (1 + sin(pi/6), pi/6) = (2, pi/6 2)
and (1 + sin(-pi/2) , -pi/2) = (0, -pi/2). This point is the origin

Answer 3

question a million y = 9x^4 - 5x² + 3x - 4 dy/dx = 36x³ - 10x + 3 question 2 y = 9x^(-a million) - 3 x^(-2) + 13 dy/dx = - 9 x^(-2) + 6x^(-3) question 3 h (x) = 6x^(-6) - 5 x^(-5) + 4 x^(a million/2) h ` (x) = - 36 x^(-7) + 25 x^(-6) + 2 x ^(-a million/2) h `(x) = - 36 /x^7 + 25 / x^6 + 2 /?x

Answer 4

There are no verticle tangents to the sin function