Spiral arc length Consider the spiral r=4θ, for θ≥0.

Question

c. Show that L′(θ)>0. Is L″(θ) positive or negative? Interpret your answer.

Accepted Answer

Welcome back, everyone. For the polar curve, are of theta equals 6 sin theta plus 8 cosine theta or theta greater than or equal to 0, determine L of theta and state whether L of theta is positive, negative, or 0. Let L of theta be equal to the integral from 0 up to theta of square root of R2 T. Plus R of T squared dt. So for this problem, we're going to identify L of data to begin with. What we want to do is identify R squared. This is our first term. We're going to take 6 sine theta plus 8 cosine theta and we're going to square it. We get 36 sin squared theta plus. 2 multiplied by 6 multiplied by 8 gives us 96. Sine, cosine theta plus 64 cosine squared theta. Now let's find the derivative, right? So What we're going to do is identify our. And this is going to be the derivative of 6. Theta + 8 cosine theta. We got 6 cosine theta minus 8 sine theta. We can now square that derivative, right? We're going to get 6 Cosine theta. Minus 8 sin theta squared which is 36. Cosine squared theta minus 96. Sine theta, cosine theta plus 64 sin squared theta. And now we can add these, right? So, R squared off T, we're going to change the variable to T. And we're going to write R2d of T plus R T squared is equal to. Notice that we have 36. Sin squared theta, which now becomes T, so we're going to write 36 sin squared T. Plus, let's group the terms with similar coefficients. So we're going to write 36 square of T plus. 36 cosine squared off T. Plus 96 st cosinet minus. 96. Sine, cosinet. Plus 64 cosine squared of T plus 64 squared of T. So we can cancel out two terms with opposite signs. 96 T cosine T. And it's opposite value with a negative sign, right? For the other chumps we can factor out. 36 in C sin squared C. Plus cosine square C. And then we're going to factor out 64, so we're adding 64 and C's sin squared of T plus cosine squared of T. According to the Pythagorean theorem, sine squared plus cosine squared is 1. So we simply get 36 + 64, which is 100. So we can now show that L of theta becomes the integral from 0 up to theta of square root of 100 because this is our simplified form of R2 plus R squad and we're integrating with respect to C. We can factor out the constant square root of 100 is 11. And we can integrate need T from 0 up to theta. We get 10 C evaluated from 0 up to theta, which is equal to 10 in print C the minus 0, according to the fundamental theorem of calculus part two. The final answer is some data. So that's L of theta, and we want to determine the L of theta, so we're going to find the derivative, right? Performing differentiation, the derivative of 10 theta. It's going to be equal to. 10 multiplied by 1, the derivative of theta is 1, and we basically get 10 for L. And now the second derivative L of theta is going to be the derivative of 10. Which is 0 because 10 is a constant. So now we can show that L is equal to. Some And L is equal to 0. Thank you for watching.

Spiral arc length Consider the spiral r=4θ, for θ≥0.

c. Show that L′(θ)>0. Is L″(θ) positive or negative? Interpret your answer.

Key Concepts

Arc Length of a Polar Curve

First Derivative of Arc Length (L′(θ))

Second Derivative of Arc Length (L″(θ)) and Its Interpretation

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