# Find Length of Parametrized Curve: x(t), y(t)

• ILoveBaseball
In summary, the length of the parametrized curve is given by the definite integral from 0 to 1 of the absolute value of the quadratic polynomial with coefficients 12, -24, and 24, which is equal to 16.
ILoveBaseball
Find the length of parametrized curve given by
$$x(t) = 0t^3 +12t^2 - 24t$$,
$$y(t) = -4t^3 +12t^2+0t$$,
where t goes from zero to one.
Hint: The speed is a quadratic polynomial with integer coefficients.

it's an arclength question right?

x' = 24*t-24
y' = -12*t^2+24*t

$$\int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2}$$

i get 6.49 when i use a math program to integrate it which is incorrect. anyone know where i went wrong?

ILoveBaseball said:
Find the length of parametrized curve given by
$$x(t) = 0t^3 +12t^2 - 24t$$,
$$y(t) = -4t^3 +12t^2+0t$$,
where t goes from zero to one.
Hint: The speed is a quadratic polynomial with integer coefficients.

it's an arclength question right?

x' = 24*t-24
y' = -12*t^2+24*t

$$\int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2}$$

i get 6.49 when i use a math program to integrate it which is incorrect. anyone know where i went wrong?

Note that:

$$1: \ \ \ \ \int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2} \, dt \ = \ \int_{0}^{1} \sqrt{ \left ( 12t^{2} \ - \ 24t \ + \ 24 \right )^{2} } \, dt \ =$$

$$2: \ \ \ \ = \ \int_{0}^{1} \left | \left ( 12t^{2} \ - \ 24t \ + \ 24 \right ) \right | \, dt \ \color{red} = \ \mathbf{\left (16 \right )}$$

~~

Last edited:

To find the length of a parametrized curve, we use the formula:

L = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} dt

In this case, a = 0 and b = 1, so we have:

L = \int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2} dt

To integrate this, we can use the substitution u = t^2, which gives us:

du = 2t dt

And our integral becomes:

L = \int_{0}^{1} \sqrt{(24*u-24)^2 + (-12*u+24*u)^2} \frac{1}{2\sqrt{u}} du

Simplifying, we get:

L = \int_{0}^{1} \frac{1}{2} \sqrt{(24*u-24)^2 + (12*u)^2} du

Now, we can use the trigonometric identity:

\sin^2 \theta + \cos^2 \theta = 1

To rewrite our integral as:

L = \int_{0}^{1} \frac{1}{2} \sqrt{24^2(u-1)^2 + 12^2(u)^2} du

= \int_{0}^{1} \frac{1}{2} \sqrt{12^2(u^2 + (u-1)^2)} du

= \int_{0}^{1} \frac{1}{2} \sqrt{12^2(2u^2 - 2u + 1)} du

= \frac{1}{2} \int_{0}^{1} 12\sqrt{2u^2 - 2u + 1} du

We can now use the substitution v = 2u^2 - 2u + 1, which gives us:

dv = (4u-2) du

And our integral becomes:

L = \frac{1}{2} \int_{0}^{1} 12\sqrt{v} \frac{dv}{4u-2}

= \frac{3}{4} \int_{0}^{1} \sqrt{v} dv

= \frac{3}{4} \left[ \

## 1. What is a parametrized curve?

A parametrized curve is a mathematical function that describes the position of a point in a two-dimensional space as a function of a parameter, typically denoted as t. This means that for each value of t, there is a corresponding point on the curve. Parametrized curves are often used to represent complex shapes and curves in a simple and efficient manner.

## 2. How do you find the length of a parametrized curve?

The length of a parametrized curve can be found using the arc length formula:
L = ∫ √(x'(t)^2 + y'(t)^2) dt
where x'(t) and y'(t) are the derivatives of the x and y components of the curve with respect to t. This formula calculates the total length of the curve by integrating the distance between each infinitesimal point along the curve.

## 3. How is the length of a parametrized curve related to calculus?

The length of a parametrized curve is related to calculus through the use of integration. Calculus allows us to find the derivative of a parametrized curve, which is then used in the arc length formula to find the total length of the curve. Integration is also used to calculate the area under the curve, which can be useful in many real-world applications.

## 4. Can you give an example of finding the length of a parametrized curve?

Sure, let's say we have the parametrized curve x(t) = cos(t), y(t) = sin(t) for t ∈ [0, 2π]. To find the length of this curve, we first calculate the derivatives: x'(t) = -sin(t) and y'(t) = cos(t). Plugging these into the arc length formula, we get:
L = ∫ √((-sin(t))^2 + (cos(t))^2) dt = ∫ √(1) dt = t + C
Evaluating the integral from t = 0 to t = 2π, we get L = 2π, which is the length of the parametrized curve.

## 5. What are some practical applications of finding the length of a parametrized curve?

Finding the length of a parametrized curve has many practical applications, such as in physics, engineering, and computer graphics. It can be used to calculate the distance traveled by an object that follows a certain path, determine the amount of material needed to create a curved structure, or create realistic and smooth animations in computer graphics. It is also used in calculus to solve optimization problems, where the goal is to find the shortest or longest path between two points on a curve.

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