Question

Find the slope of the curve r=cos(2theta) at theta=pi/6.

Must this be solved parametrically? If so, why?

Answer 1

Find the slope of the curve\[r=\cos 2\theta\]at\[\theta= pi/3\]Must this be solved parametrically? If so, why?

Answer 2

If by parametrically, you mean let x=r cos(theta), y=r sin(theta), then getting dy/dx, then yes. That's because dy/dx represents the slope of the tangent line through a particular point on the graph.

Answer 3

I would say:
slope is a ratio of change of the dependent variable over the independent. If you have an implicit form of the function, bouth variables are treated the same way (bouth independent). Parametricly, you could express one as function of the other (some conditions must hold). In your case parametric equation could be:
\[Curve = (\theta,r(\theta))\]
so so called slope would be:
\[r(\theta) \over \theta\]

Answer 4

But why could I not take the derivative of r with respect to theta, and then substitute in pi/3 to find the requested slope?

Answer 5

You might want to look at this: http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx

Answer 6

I still don't understand. Suppose the equation was\[r=\cos(2x)\]and I was told to find the slope of r at\[x=pi/6\]Then the answer would be\[dr/dx=-2\sin(2x)\]and this equation evaluated at pi/6 equals -sqrt(3).

What is different about the problem I mentioned to start with?

Answer 7

You want the Cartesian slope of a polar curve so you need to parametrize the polar curve in terms of Cartesian coordinates x and y.

Answer 8

Is it true that the only indication that this is a polar curve is the use of the variables r and theta in the equation?

In other words, if the equation was y=cos(2x), then is it true that this would not have to be a polar curve?

Answer 9

That's right, we are talking about conventional use of symbols here.
Most would take the curve you gave as being polar.

Answer 10

Ah, many thanks!

Answer 11

ur welcome.