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The Length of Curve Calculator finds the arc length of the curve of the given interval. The curve length can be of various types like Explicit, Parameterized, Polar, or Vector curve.
**What is the Length of the Curve?**

**“The length of the curve is used to find the total distance covered by an object from a point to another point during a time interval [a,b]”**
The length of the curve is also known to be the arc length of the function.
Consider a function y=f(x) = x^2 the limit of the function y=f(x) of points [4,2].
**Where:**
**How to Find the Length of the Curv****e?**

For finding the Length of Curve of the function we need to follow the steps:
**Explicit Curve y = f(x):**

Consider a graph of a function y=f(x) from x=a to x=b then we can find the Length of the Curve given below:
$$ \hbox{ arc length}=\int_a^b\;\sqrt{1+\left({dy\over dx}\right)^2}\;dx $$
**Parameterized Function**:

If the curve is parameterized by two functions “x” and “y”. You can find the double integral in the x,y plane pr in the cartesian plane.
**Where:**
**x=f(t), and y=f(t) **
**The parameter “t” goes from “a” to “b”.**
Then the formula for the length of the Curve of parameterized function is given below:
$$ \hbox{ arc length}=\int_a^b\;\sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2}\;dt $$
It is necessary to find exact arc length of curve calculator to compute the length of a curve in 2-dimensional and 3-dimensional plan
**The Polar Function:**

Consider a polar function r=r(t), the limit of the “t” from the limit “a” to “b”
$$ L = \int_a^b \sqrt{\left(r\left(t\right)\right)^2+ \left(r'\left(t\right)\right)^2}dt $$
In mathematics, the polar coordinate system is a two-dimensional coordinate system and has a reference point. The distance between the two-point is determined with respect to the reference point. It can be quite handy to find a length of polar curve calculator to make the measurement easy and fast.
**The Vector Values Curve:**

The vector values curve is going to change in three dimensions changing the x-axis, y-axis, and z-axis and the limit of the parameter has an effect on the three-dimensional plane. You can find triple integrals in the 3-dimensional plane or in space by the length of a curve calculator.
The formula of the Vector values curve:
$$ L = \int_a^b \sqrt{\left(x'\left(t\right)\right)^2+ \left(y'\left(t\right)\right)^2 + \left(z'\left(t\right)\right)^2}dt $$
**Example:**

Find the length of the curve of the vector values function x=17t^3+15t^2-13t+10, y=19t^3+2t^2-9t+11, and z=6t^3+7t^2-7t+10, the upper limit is “2” and the lower limit is “5”.
Given:
Lower limit= 5, upper limit = 2
Sol:
The length of the curve is given by:
$$ L = \int_a^b \sqrt{\left(x'\left(t\right)\right)^2+ \left(y'\left(t\right)\right)^2 + \left(z'\left(t\right)\right)^2}dt $$
First, find the derivative x=17t^3+15t^2-13t+10
$$ x '\left(t\right)=(17 t^{3} + 15 t^{2} - 13 t + 10)'=51 t^{2} + 30 t - 13 $$
Then find the derivative of y=19t^3+2t^2-9t+11
$$ y '\left(t\right)=(19 t^{3} + 2 t^{2} - 9 t + 11)'=57 t^{2} + 4 t - 9 $$
At last, find the derivative of z=6t^3+7t^2-7t+10
$$ z '\left(t\right)=(6 t^{3} + 7 t^{2} - 7 t + 10)'=18 t^{2} + 14 t - 7 $$
Finally, calculate the integral:
$$ L = \int_{5}^{2} \sqrt{\left(51 t^{2} + 30 t - 13\right)^2+\left(57 t^{2} + 4 t - 9\right)^2+\left(18 t^{2} + 14 t - 7\right)^2}dt $$
**How does the Length of Curve Calculator Work?**

You just stick to the given steps, then find exact length of curve calculator measures the precise result.
**Input:**
**Output:**
**References:**

From the source of Wikipedia: Polar coordinate,Uniqueness of polar coordinates
From the source of tutorial.math.lamar.edu: Arc Length, Arc Length Formula(s)

**Points [4,2]= limit of the function,****The upper limit = 4****The lower limit = 2**

- First, find the derivative of the function,
- Second measure the integral at the upper and lower limit of the function.

- Choose the type of length of the curve function
- Enter the function
- Write the upper and lower limit
- Hit the calculate button

- Length of the curve

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