I'm sorry, I cannot answer this question as it is not a topical webpage title.

In summary, the problem is to find the length of the curve y=x^3 using the starting point P(1,1). The relevant equation is f(x)=\int_{a}^{x} {\sqrt 1 + {f'(t)}^2}, and the initial attempt at a solution involved finding the y' and plugging it in, resulting in 1+9t^4 inside the square root. However, this requires the Elliptic Integral of the First kind, which is not a simple trig sub. Another possible solution is to use the function L(x) = \int_1^x \sqrt{1 + 9t^4}dt, which represents the length along the curve and can
  • #1
MillerGenuine
64
0

Homework Statement



Find the length of the curve y=x^3 using P(1,1) as the starting point

Homework Equations




[tex]
f(x) = \int_{a}^{x} {\sqrt 1 + {f'(t)}^2}
[/tex]



The Attempt at a Solution



So far all I've done is found my y' and plugged it in, giving me 1+9t^4 inside the square root...now from here I am not sure whate technique to use to do the integral.
 
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  • #2
looks like a classic trig sub integral
 
  • #3
lanedance said:
looks like a classic trig sub integral

If only. Checking Wolfram-Alpha, you'll require the Elliptic Integral of the First kind to solve this. A simple trig sub works with functions of the form sqrt(a^2 + (bx)^2), but not sqrt(a^4 + (bx)^4).
 
  • #4
fair bump, play on
 
  • #5
MillerGenuine said:

Homework Statement



Find the length of the curve y=x^3 using P(1,1) as the starting point

Homework Equations




[tex]
f(x) = \int_{a}^{x} {\sqrt 1 + {f'(t)}^2}
[/tex]



The Attempt at a Solution



So far all I've done is found my y' and plugged it in, giving me 1+9t^4 inside the square root...now from here I am not sure whate technique to use to do the integral.
Is there an end point given? If so, you run into the problem pointed out by Char Limit. If not, are you just supposed to find an expression that represents the length from (1, 1) to an arbitrary point (x, x3)?

If it's the latter, here's a function that gives the length along the curve. Although difficult to integrate analytically, it can be approximated by a number of numerical integration techniques.
[tex]L(x) = \int_1^x \sqrt{1 + 9t^4}dt[/tex]
 

1. What is the Arc length Function in Calculus II?

The Arc length Function, also known as the Length of Curve Function, is a mathematical concept that calculates the length of a curve on a graph. It is commonly used in Calculus II to find the distance traveled by a moving object along a curved path.

2. How is the Arc length Function calculated?

The Arc length Function is calculated by integrating the square root of the sum of the squares of the derivatives of the curve's x and y components. This results in a single function that represents the length of the curve from a starting point to an ending point.

3. What is the significance of the Arc length Function in Calculus II?

The Arc length Function is an important concept in Calculus II as it allows for the calculation of distances along curved paths, which cannot be easily measured using traditional methods. It also has practical applications in physics and engineering, where it can help determine the length of a curved trajectory or the distance traveled by an object along a curved path.

4. Can the Arc length Function only be used for 2-dimensional curves?

No, the Arc length Function can also be used for 3-dimensional curves, where it is known as the Space Curve Arc length Function. In this case, the function is calculated by integrating the square root of the sum of the squares of the derivatives of the curve's x, y, and z components.

5. Are there any limitations to using the Arc length Function?

The Arc length Function has some limitations, as it can be difficult to calculate for certain types of curves, such as spirals or curves with infinite length. In these cases, numerical methods may be used to approximate the arc length. Additionally, the function may not be applicable to curves that are not differentiable or have sharp corners.

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