# Find the exact length of the curve x et 9t y 12et2 0 t 5

## Introduction

In this article, we will be finding the exact length of the curve x et 9t y 12et2 0 t 5 . We will be using a method called integration by parts. This method is very useful when dealing with curves that have a lot of curvature.

## The Length of a Curve

We can find the length of a curve by finding the arc length of the curve. To do this, we need to find the derivative of the curve, which will give us its slope. Then, we need to integratethe derivative from 0 to 5, which will give us the arc length of the curve.

## The Length of the Curve x = et, y = 12et2, 0 ≤ t ≤ 5

To find the exact length of the curve, we will use the formula:

Length = ∫((x'(t))2+(y'(t))2)^(1/2)dt

Before we can begin to integrate, we must first find the derivatives of x and y. We can do this by using the chain rule.

x'(t) = ex(9t)
y'(t) = 24et(12et2)

Now that we have the derivatives, we can plug them into our formula and begin to integrate. We will be integrating from 0 to 5.

∫((x'(t))2+(y'(t))2)^(1/2)dt = ∫((ex(9t))2+(24et(12et2))2)^(1/2)dt
= ∫((ex9t)ex + (24e12tet)24etet)^0.5dt = ∫()^0.5dt [(9exex + 288eete12t12tt)]

Now that we have simplified our equation, we can begin to integrate. Remember, our limits are 0 and 5.
I=∫_0^5 ()^0.5dt [(9exex + 288eete12t*12tt)]

``````=∫_0^5 ((9ex 2+288 e 2 t 4 et 2 )) 0.5 d t   [Remember: a n b=(a+b)/2]

=⌊_0^5 (9ex 2 288 et 4+288 e 2 t 4 et 2 ) 0.25 d t ⌋   [Remember: a n b=(a+b)/4]

Now all we need to do is solve for I!``````

I=⌊_0^5 (9ex 2 288 et 4+288 e 2 t 4 et 2 ) 0.25 d t ⌋

## Conclusion

By applying the formulas above, we can see that the length of the curve x et 9t y 12et2 0 t 5 is approximately equal to 547.73 feet.