Find the length of the curve below on the interval [ 0 , 4 ] \left\lbrack0,4\right\rbrack . x = 1 3 ( 2 t + 3 ) 3 2 x=\frac13\left(2t+3\right)^{\frac32} , y = 1 2 t 2 + t y=\frac12t^2+t

So in this problem, we're asked to find the length of the curve below on the interval from zero to four. And you can see that we're given this x as a function of t, one third times two t plus three to the three halves power, and we're given y as a function of t, one half t squared plus t. So it's clear that we're dealing with these parametrics, and since we're trying to find the length of the curve, that means we're going to be dealing with an arc length here. Now the formula for the arc length looks like this. It's going to be the integral from a to b of the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared, and all of this is integrated with respect to t. So what I need to do is figure out how to properly set up this integral. If I can set this up, then I just have to go through the process of solving the integral, and that will get me to my solution. So let's see how we can do this. Well, first off, I'm going to find my dx over dt, and that's simply going to be the derivative of x, which we have right here with respect to t. Now this derivative is going to be a little bit tricky because we have a lot going on as far as this function goes. But what I can do is I can use the chain rule here. So you take the derivative of the outside first, then multiply that by the derivative of the inside. So starting with the outside function, we can just use a simple power rule. Now doing this is going to give us three halves times one third, and we'll keep the inside the same, so that'll be two t plus three. And then we have three halves minus one. Well, three halves is the same thing. Minus one is the same thing as minus three over three, which of the or we could say the same thing as two over two, that because we want the same denominator here, and that's going to give us one half. So that gives us one half right about here. Now keep in mind, this is just the derivative of the outside. We need to multiply this now by the derivative of the inside function using the chain rule, and the derivative of two t plus three with respect to t, all of that just comes out to two, because the t will come off and then the derivative of any constant is zero. Now there is some simplification I can do to make things look a little better. I can cancel the three that we have right here, and I can also cancel the two that we have up here. So that means all we're going to be left with is 2t plus three to the one half power, which is the same thing as the square root of 2t plus three. So this is our derivative of x with respect to t. Now next I need to deal with our derivative of y with respect to t. Well-to-do this, what I'm going to do is take the derivative of y. Now I can go ahead and use the power rule on this first term, and that's going to give us two times one half t to the two minus one, which is just one. Then we would have plus, and then it's going to be the derivative of t, which is just one. This is all the derivative taken with respect to t. Now here, I can cancel the two with that one right there, giving me that this whole thing is just equal to t plus one. So now we have our dy over dt and our dx over dt, which means we can plug it into the arc length equation we have right here for this integral. So we're going to have the integral, and a to b is the interval that we're interested in, which you can see the interval we're given is from zero to four. So we're gonna go from zero all the way up to four. We're going to have the square root of dx over dt squared. Now dx over dt, I can see that that's just the square root of 2t plus three. This is going to be the square root of 2t plus three, and this whole thing is squared. And what I'm going to do is add on the derivative of y with respect to t squared. To do the derivative of y with respect to t is t plus one. And then this is all integrated with respect to t. So now I have this integral properly set up. So now I just need to go through the process of solving this. But we're gonna keep these the same. The integral is gonna stay the same. And then what I can do is cancel this square with the square root right there. This is gonna leave me with two t plus three. And then for this t plus one, well, I can recognize that's just the same thing as plus t plus one squared. That's going to be all with respect to t. Now I think that there's actually more I can do to simplify here, but it's gonna be best if I try to foil out this t plus one. So we have t plus one squared, which is the same thing as t plus one times t plus one. T times t is t squared, then we're gonna have plus t plus one times t, which is t, and then we're gonna have plus one times one, which is one. So this will all give us t squared plus two t plus one. So that's what this whole expression here is going to be equal to. So what that means is I can rewrite this whole thing as the integral from zero to four with the square root, and we're gonna have two t plus three plus t squared plus two t plus one. Now let's see how I can make this even more simple. Well, what I can see right here is that we have a two t and a two t right there we can combine to give us four t. And actually, I'm gonna put it like this. I'm gonna write the t squared first, and then we're going to have the two t plus two t, which is four t, and then I'm going to add the three and the one, which three plus one is four. So it's still gonna be integrated with respect to t. Now what can I do at this point? Well, what I can do is recognize something. We have t squared plus four t plus four. This actually looks like something that we've seen before. This t squared plus four t plus four, that's actually the same thing as t plus two squared. And think about it. Because if we had t plus two times t plus two, we could actually foil this all out, and we would get t squared plus two t plus two t plus two times two, which is four. And all of that is the same thing as exactly what we have inside the the radical right there. So what that means is I can write this whole thing as the integral from zero to four of the square root of t plus two squared. And in this case, the square and square root are going to cancel each other, giving us the integral of t plus two. The integral from zero to four of t plus two. And this is actually going to be the absolute value of t plus two because when you square something and then take the square root, you can only get positive results that come from this. But for all intents and purposes of evaluating this integral, we just evaluate each one of these separately, recognizing that when we go from zero to four, this will always give us a positive value anyways. So doing this, what we're going to do is first integrate this t. And I can use the antiderivative or the power rule, which is gonna give me t squared over two. Then Then I can go ahead and take the antiderivative of two, which is gonna give me two t, and all of this is going to be bounded from zero to four. Now at this point, what I need to do is plug in my balance. So we're gonna go ahead and plug in this four that we have right here, the upper bound, which is gonna give me four squared over two plus two times four. And then I'm going to go ahead and take this whole thing and subtract off the lower bound plugged in. Plugging in the lower bound is going to give us zero squared over two plus two times zero. So this whole thing really just comes out to zero. And if you go ahead and crunch these numbers right here, you can go ahead and do this on a calculator or by hand, All of this should come out to 16. So 16 is going to be the arc length, s, and that would be the solution to our problem. Hope you found this helpful.

16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

Calculus

16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

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Multiple Choice

Find the length of the curve below on the interval $open bracket 0 comma 4 close bracket$ .
$x = \frac{1}{3} {(2 t + 3)}^{\frac{3}{2}}$ , $y = \frac{1}{2} t^{2} + t$

$16$

$24$

$12$

$20$

Verified step by step guidance

Step 1: Recall the formula for the length of a curve in parametric form. If the curve is defined by x = f(t) and y = g(t), the length of the curve on the interval [a, b] is given by:

\int_{a}^{b} \sqrt{{\frac{dx}{dt}}^{2} + {\frac{dy}{dt}}^{2}} dt

Step 2: Compute the derivatives dx/dt and dy/dt. For x = (1/3)(2t + 3)^(3/2), use the chain rule to find dx/dt. For y = (1/2)t^2 + t, use the power rule to find dy/dt.

Step 3: Substitute dx/dt and dy/dt into the formula for the curve length. This will involve squaring both derivatives and adding them under the square root.

Step 4: Set up the integral for the curve length over the interval [0, 4]. The integral will be of the form:

\int_{0}^{4} \sqrt{{\frac{dx}{dt}}^{2} + {\frac{dy}{dt}}^{2}} dt

Step 5: Evaluate the integral either analytically or numerically to find the length of the curve. This step involves solving the integral, but the final numerical result is not calculated here.