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Numerade Educator



Problem 2 Easy Difficulty

Find the length of the curve.
$$\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+\ln \cos t \mathbf{k}, \quad 0 \leqslant t \leqslant \pi / 4$$


\ln |\sqrt{2}+1|


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Video Transcript

we went to evaluate the length of the curve, as described by the vector function are of tea which could be written as the co sign T defector I waas sine of t time is the back to the J plus the natural log of the two sine of t was the vector k and this is on the interval, zero to pi over four. So what's first? By taking the derivative of R factor function. So we have our prime of tea which, if we, uh, evaluate the first, the derivative of the first component function, we should get negative sign of tea. Plus the derivative of sine of t is co sign of tea and we have to use the chain role to evaluate the third component function. So what we'll get is one over co sign of tea times negative sign of tea. And if you want to simplify about a little bit, we can simplify that last term as negative tangent of t since sign of her co sign is equal to tangent. All right, so we have our derivative. Next thing we want to do is find the magnitude of that resulting vector function so the magnitude of our prime of tea Take the square root of our first component function. Negative sign of t squared plus co sign of tea Well squared plus negative tangent of t quantity squared and simplifying that a little bit We end up with sine squared of teeth plus co sine squared of tea plus hand agent squared T we can recognize are the thuggery and identity is one of our trig identities Sine squared plus co sine squared is one So we end up with one plus tangent square toe t and we can see another one of our factory and identities Ah, one plus tangent Square T is gonna be seeking squared of tea So we end up with a square root of Seacon Square t which is equal Teoh Sequent of do you? So now we're in a good position to be able Teoh, evaluate the length of our curve. So we have l equals the integral from zero to pi. Over four was the interval. We were given, uh, seek it square or sigint of tea. Do you t and the integral of second of T is the natural log Absolute value of sequent t plus tangent of T. I'm gonna use the fundamental theorem of calculus to evaluate that from zero to pi over four. Sir, what we get is natural long, uh, seeking to pi over for plus tangent pi over four minus natural log Sequent of zero plus Tangent of Zero which we can simplify by evaluating those trick functions. So, seeking a pi over four squared of to tangent of pi over four is one minus natural log. Uh, seeking of zero is one and tangent of 00 So this, uh, last term ends up being natural log of one which we know is zero. So we can cross that out and we end up with the natural log, uh, squared of two plus one, which represents the length of the curve, as described by the vector function, are of tea.

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