Verified answer

To find the slope of the tangent line we need to find dy/dx.

(In the following calculations I will replace theta with A for sake of brevity.)

Now y = r*sin(A) and x = r*cos(A), and we are given that r is a function

of A such that r(A) = 1 + sin(A). The chain rule gives us dy/dx = (dy/dA) / (dx/dA),

so we need to calculate both dx/dA and dy/dA.

Now dx/dA = (dr/dA)*cos(A) - r*sin(A) = cos(A)*cos(A) - (1 + sin(A))*sin(A) =

cos^2(A) - sin^2(A) - sin(A) = cos(2*A) - sin(A).

Next, dy/dA = (dr/dA)*sin(A) + r*cos(A) = cos(A)*sin(A) + (1 + sin(A))*cos(A) =

2*sin(A)*cos(A) + cos(A) = sin(2*A) + cos(A).

So at A = 0.4 we have dy/dx = (dy/dA) / (dx/dA) =

[sin(0.8) + cos(0.4)] / [cos(0.8) - sin(0.4)] = 5.332 to 3 decimal places.

The trick to this question is to not plug in for θ right away.

First, calculate the general function dy/dθ. Do you know how to take derivatives?

If so, once you have dy/dθ, you will get a function of θ. (Hint: it will be exactly one trigonometric term depending on θ.) Plug in for θ=0.4.

Now you have a value for dy/dθ at θ=0.4. The value returned, since it represents the derivative at that point, equals the slope of the tangent line of the curve at that point.

So both questions you ask (about dy/dθ and the slope of the tangent line) have the same answer.