Tangents and normals: Let a polar curve be described by r = f(...

Question

Tangents and normals: Let a polar curve be described by r = f(θ), and let ℓ be the line tangent to the curve at the point P(x,y) = P(r,θ) (see figure).

e. Prove that the values of θ for which ℓ is parallel to the y-axis satisfy tan θ = f(θ)/f'(θ).

Graph of a polar curve r = f(θ) with tangent line ℓ at point P(r, θ), showing angles θ, φ, and α.

Accepted Answer

Welcome back everyone. Find the length of the polar curve R of theta equals 2 cos theta from theta of pi divided by 4 to theta of pi divided by 3. For this problem, let's begin with the Euler RL formula. L is equal to the integral from alpha to beta. Of square root of R squared plus. The derivative of R with respect to theta squared the theta. So what we want to do is identify each parameter. We can start with the limits of integration. We're integrating from alpha of pi divided by 4 up to beta of pi divided by 3. Now we're going to find R squared. Specifically, we want to square the function itself, so that'd be 2 cosequant theta squared, which is equal to 4 cosequant squared of theta. Now let's find the derivative of R with respect to theta. That's TR divided by the theta. We want to differentiate two cosequences. Sarah And we are going to get. 2, which is a constant we can factor it out, right? And now the derivative of cosequent is negative cosequent cotangent. So we can write -2 cosequent theta cotangent theta. And we can square it. The derivative of R with respect to theta squared is -2 cos. Theta cotangent theta squared. Which is 4 cosican squared of theta. Multiplied by cotension squared of theta. Not that we want to add these two squares, right? So basically R squared plus. The derivative of R. Squared is going to be equal 2. For Cosik and squared of the Plus or cosican squared of theta. Cotension squared of theta. We can factor it out. We can seek out 4 cos squared of theta, which leaves us with 1 plus cotension squared of theta reiss. And now let's recall that a 1 plus cotention squared is coc squared. So we get 4 co squad of theta. Multiplied by coc squared of theta. It can be simplified into. 4 cosequent to the power of 4 of Thea. Now notice that we're taking square root of that term, so we can now show that square root of R squared plus the derivative of R with respect to theta squared. Is equal to square root of For co sequence to the power of 4 data. And it is going to be equal to 2 cosequent squared of theta. So now we have our integrant and we can integrate L. Our our client is going to be the integral from negative, I'm sorry, from divided by 4. Up to pi divided by 3. Off to Cosican squared of the di theta. Now, we can factor out 2, and we can integrate cos squared. The integral of cos theta is negative cotangent theta. So we get -2 cotangent theta. Evaluated from 5 divided by 4. Up to pi divided by 3. And now let's calculate. The result Once again we can factor out -2. And take cotangent of pi divided by 3 minus cotangent of pi divided by 4. This is equal to -2 multiplied by. In parenthesis square root of 3 divided by 3 minus 1. We can now use the common denominator, which is 3. -2 multiplied by. -1 is 2 and 2 multiplied by 3 gives us 6. And we are going to subtract 2 multiplied by a square root of 3, right? So 2 square root of 3. Therefore, the final answer to this problem is 6 minus 2 square root of 3 divided by 3 units. Thank you for watching.

Tangents and normals: Let a polar curve be described by r = f(θ), and let ℓ be the line tangent to the curve at the point P(x,y) = P(r,θ) (see figure).
e. Prove that the values of θ for which ℓ is parallel to the y-axis satisfy tan θ = f(θ)/f'(θ).

Key Concepts

Polar Coordinates and Conversion to Cartesian Coordinates

Slope of the Tangent Line to a Polar Curve

Condition for a Line to be Parallel to the y-axis

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