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Problem 37 Hard Difficulty

Use the formula in Exercise 36 to find the curvature.
$$x=e^{t} \cos t, \quad y=e^{t} \sin t$$

Answer

$$
\kappa(t)=\frac{1}{\sqrt{2} e^{t}}
$$

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Video Transcript

in this problem, we want to use the resulting formula for exercise 36 which is in the top right corner to find the curvature of the function that's represented by the component functions X equals eat of the power tee times CO sign of tea and why equals even the power of t times sine of t. So what we want to do is take the first derivative and the second derivative of each of these two component functions. So let's focus on X first, so x with the dot over top, which is the first derivative of X is going to be the horny is a product rule here so derivative of the first, uh, part of T is going to be e t times the second function, which is co sign of tea, plus the first function of t times the derivative of CO sign of T. Which is gonna be negative. Sign of t. So I'm actually gonna change my sign here. There we go. Now, if you want to take the second derivative, we're gonna have to use our product rule twice here, so derivative of the first, um, using the product rule, we get e to the power of T Times Co Sign of tea minus eat in power of tea time. Sign of to you which actually is the first derivative. I have the entire function and then we're going to subtract the derivative of the second function again using the product rule so derivative of e of t just e of t Multiply that by the second function sine of t When we add the first function E of t times derivative of the second function which is co sign of and I'm actually a simplify each of those just a little bit So first derivative of X on a factor out each of the power of teeth. So we have co sign of tea. My sign of t is the second factor. And then let's fix up our second derivative a little bit so e to the power of T Times co sign of tea. Yes, even the power tee times sign of t we're going to distribute or negative sign here and then add up are like terms. So are, uh, co sign terms will add up to be zero and then we're just left with negative two times e to the power of T times sine of t as our second derivative of X. Now let's go ahead and find the first and second derivatives of why so first derivative of why, using our product rule, we end up getting each of the power of t times sine of t plus. You have the power of T Times co sanity, uh, and then for our second derivative of why we end up with either the power of t times sine of t plus eat of the power of T Times Co sign of T. And that's applying the product routes to the first term. One for the second term will have either the power of t times co sign of tea minus eat of the power of t times sine of t So simplifying that adding up our like terms are signed terms we're going to add up to be zero. So we're left with two times e to the power of T co sign of tea and then, for my first derivative, I'm going Teoh factor out e to the power of T just like I did with the first derivative of X. All right, so now we went Teoh, use the formula in the top right corner for capital of tea. So what we're gonna have is in our numerator first derivative of X time. Second derivative of why? Just to eat of the power of T co sign of tea minus first derivative of why U to the power of T sign of t of plus co sign of tea time second derivative of X which is negative to eat of the power T time science t worry a simple vibius. And just a minute here, an inner denominator. We're going to have first derivative of X squared. So it's gonna give us e to the power of to t times care sine of t minus sine of t quantity squared on first derivative of why squared is going to be e of the power of to t times sine of t plus hair sine of t squared on. That denominator is going to be raised to three house Power Mary. So let's simplify this a bit In our numerator, we're going to have absolute value to times eat of the power of to t times her sign squared of tea minus sine of t times co sign of tea plus two times eat of the power to t time sine squared of tea plus sign t co sign T than in our denominator. We're going to have e to the power of to t. Then we'll have for co sign of coastline of T minus sine of t all squared. We're gonna end up a co sine squared of tea minus, Let's see, minus two sine of t co sign of tea plus signs weird of tea. And then the second term, we're gonna hav e to the power of to T and then sign of t plus co sign of t quantity squared is going to be signed squared of tea plus two sine of t co sign of tea plus her sine squared of tea and that entire denominators raised to the three house power again, Right? Let's continue our simplification here. So I'm gonna do in my numerator is factor out two times e to the power of to t and then all of these terms, we're going to be added together and we can see that sign tee Times Co sign of tea and negative scientific co sign if he will add to be zero. So we're just left with Kerr sine squared of tea plus sine squared of tea and our denominator here we're also going to factor out the term e to the power of to t You've got right here which results on all of these terms being added together. So are negative to sign if he co sign if t positive to sign of tee times co sign of tea will add up to be zero So and we end up with coastline squared of T plus sine squared of tea Quest sine squared of tea plus Carson squared of tea Well, that's raised to three halfs power when we see some nice things happening here because we have co sign a squared of TP plus sine squared of tea which is just equal toe one. So our numerator we've got absolute value of two times e to the power of to t, which is always positive that or what value of tea we plug in so we don't have to read our absolute value symbols. And then in our denominator, we have to sine squared of T plus coast on squared of t. So these equal one you got One plus one is two times U to the power of to t all raised to the three halfs power. And if we want, we can simplify even a little bit further or so. What we'll get is one over. I swear to to times e to the power of T. And this represents the curvature of the functions represented by the ex and why we were given.

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