United States Calculus 2 - Calculus in Polar Coordinates

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Homework Statement

Find the slope of the line tangent to the polar curve at the given point. At the point where the curve intersects the origin, find the equation of the tangent line in polar coordinates.

r = 6 sinθ; (-3 7∏/6)

Homework Equations

The Attempt at a Solution

Select the correct choice below and, if necessary, fill in the answer box within your choice.

A. The slope of the curve at the point (-3, 7∏/6) is _
B. The slope of the curve at the point (-3, 7∏/6) is undefined.

I selected A. and put sqrt(3) and it told me I was correct.

The equation of the tangent line when the curve intersects the origin is _
(Type an equation. Type any angles in radians between 0 and 2∏.)

I'm not sure how to do this part. I believe I put in the correct equation in Cartesian. coordinate system in terms of y and x and it told me I was wrong. I wasn't exactly sure why and then I realized that it wanted me to input a polar coordinate equation in the instructions, which I'm not exactly sure how to do.
When I put the equation it told me I was wrong and this came up

If the graph of the polar curve passes through the origin of some angles θ_0, then f(θ_0)=0 and dy/dx=tan θ_0, which means that the equation of a tangent line through the origin is θ=θ_0. Thus, to determine the equation of the tangent line(s) when the curve intersects the origin, find the value(S) of θ when r=0. Note that this hold for f'(θ_0)=/=0.

I'm not exactly sure what it wants me to do.

I got
r= 6sinθ
x=6sinθcosθ
y=6(sinθ)^2
dy/dx= ( 2cosθsinθ )/( (cosθ)^2 - (sinθ)^2 )

I'm not exactly sure why they tell me dy/dx = tanθ_0 or why the equation of the line that it wants me to enter is just an angle θ=θ_0 if that is what it wants me to do.

I do not see what setting r=0 and solving for θ does

r = 6sinθ = 0
θ = 0,∏,2∏

Is this the answer it wants?

Thanks for any help.

 

[mighty2000]

Thank you for your post. I can see that you have made some good progress in finding the slope of the tangent line for the given polar curve and point. However, there are a few key concepts that you seem to be missing in your attempt at finding the equation of the tangent line in polar coordinates.

Firstly, when finding the slope of a tangent line to a polar curve, we use the formula dy/dx = (dr/dθ) / (rdθ/dθ). This is because we are dealing with polar coordinates, where the independent variable is θ and not x. So we need to use the chain rule to find the derivative of r with respect to θ, and then divide it by the derivative of θ with respect to itself, which is just 1.

In your attempt, you have used the formula dy/dx = (2cosθsinθ) / ((cosθ)^2 - (sinθ)^2). This is incorrect because you have not used the correct derivative for r with respect to θ. The correct derivative is dr/dθ = 6cosθ.

So the correct formula for the slope of the tangent line in this case would be dy/dx = (6cosθ) / (6cosθdθ/dθ). Simplifying this, we get dy/dx = (6cosθ) / (6cosθ). This simplifies to dy/dx = 1, which is a constant value and does not depend on θ.

This leads to the second key concept that you have missed. When the slope of a tangent line is a constant value, it means that the tangent line is a horizontal line. This means that the tangent line will have an equation of the form y = constant. In this case, since we are working in polar coordinates, the equation of the tangent line will be r = constant.

Now, when we are looking for the equation of the tangent line at the origin, we need to find the value of θ at which r = 0. This is because the origin is where the curve intersects the x-axis, which is the line where r = 0. So we need to solve the equation r = 0 for θ.

r = 0
6sinθ = 0
sinθ = 0
θ = 0, ∏, 2∏

As you have correctly found, the values of θ at