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# Find the length of the curve.$$\mathbf{r}(t)=12 t \mathbf{i}+8 t^{3 / 2} \mathbf{j}+3 t^{2} \mathbf{k}, \quad 0 \leqslant t \leqslant 1$$

## $$|15|$$

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all right. In this problem, we want to evaluate the length of the curve, as described by the vector function are of tea, which we have written as 12 tee times. Doctor I plus eight t to the three halves Power Inspector J plus three t squared times, vector K. And that's on the interval from 0 to 1. So let's start by taking the derivative of our vector function with respect to t. So we have our prime tea and the derivative of 12 to you with respect to T is going to be 12 keeper vector component I If we evaluate eight times T to the three halfs power that derivative with respect to T, what we're gonna get is eight times three halves t to the 1/2 power. I'm Inspector J. Plus, that last derivative is going to be six t respect, okay. And just simplifying that middle term in particular, we will have times 3 12 times t to the 1/2 power. We'll carry the rest down. All right, so here we have our derivative. Let's go ahead and take the magnitude of our prime of tea, which we can rate is the square root, uh, 12 squared plus 12 t to the 1/2 squared plus six t squared, which ends up being 144 plus 144 t plus 36 t squared. And we can recognize that, um, as 12 plus six t squared, which helps us out a little bit. And because that 12 plus 60 is always going to be positive, we can. Since Tia's positive, we can take the positive square roots. We end up with 12 plus six t as that value. So now that we have the magnitude of the derivative of R F T, you can evaluate the length of the curve on interval 0 to 1 12 plus six t do you tear and we end up with 12 t plus three t squared. Using her fundamental theorem with calculus, we end up with 12 times one plus three times one squared minus 12 times zero plus three times zero squared. And that last term second term is gonna all be zero. Which means that our remaining term is 12 plus three or 15 which equals the length of the curve is described by our of tea

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