Verified answer

First, remember that x = rcos(θ) and y = rsin(θ)!!!!!!!

x = 4sin(θ)cos(θ)

y = 4sin²(θ)

For polar functions, to find dy/dx, you do this:

dy/dx = dy/dθ / dx/dθ

dy/dθ = 8cos(θ)sin(θ)

dx/dθ = 4cos(2θ)

dy/dθ / dx/dθ = 8cos(θ)sin(θ)/4cos(2θ)

= 2cos(θ)sin(θ)/cos(2θ)

Now, just plug in θ = π/6:

8cos(π/6)sin(π/6)/4cos(π/3)

= √3

√3 is the same as 3^(1/2), so the answer is 'd'

The question is stated more than a little ambiguously, but I presume you mean:

(i) the polar curve is given by : r = f(Θ) = 4 sin Θ.

[Without any mention of ' r ,' it's NOT a polar equation! It simply defines a function f(Θ) of Θ --- no big deal, but that's NOTHING to do with a polar equation. I'm afraid that the previous two responders just didn't recognize that.]

(ii) I also presume that the slope of the tangent is required with respect to the line Θ = 0, and NOT with respect to the polar curve itself.

In that case, the answer is d.) 3^(1/2). Here's why:

Let the tangent make an angle ' fi ' with respect to the radial vector. Then standard results (easily seen from a simple sketch) are:

(a) tan fi = r dΘ / dr ; and

(b) the angle the tangent makes with the line Θ = 0 is fi + Θ.

Now for the calculation:

If r = 4 sin Θ, then dr / dΘ = 4 cos Θ, therefore

tan fi = (4 sin Θ) / (4 cos Θ) = tan Θ (!)

Therefore fi = Θ = π/6. Thus the angle the tangent makes with the line Θ = 0 is fi + Θ = π/6 + π/6 = π/3. And tan π/3 = 3^(1/2), of course.

Hence the answer is d.) 3^(1/2)

Live long and prosper.

You already calculated your spinoff. Now, plug on your factor (0,0) and prepare that it yields a million/2, it is the slope of your tangent. Now, set the tangent equivalent on your function, i.e. x/2 = (2x)/... remedy this for x. Your ideas are then the x coordinates of the factors which remedy the two equations, i.e. the intersects between the two.

f'(Θ) = 4cosΘ

f'(π/6) = 4cos(π/6) = 2√3

f'(Θ)=4cosΘ

f'(π/6)=4cos(π/6)=4*3^(1/2)*1/2=2*3^(1/2)