A curve is given by the following parametric equations. x(t)=t^2sin(t), y(t)=t^2cod(t). Find the arc length, S, of the curve over the interval 0 is less than or equal to t, is less than or equal to 4. Enter the exact value without decimals. S= _______ units 

To find the arc length \( S \) of a parametric curve defined by \( x(t) \) and \( y(t) \) over an interval \( a \leq t \leq b \), we use the formula:

\[
S = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt
\]

Given:
\[
x(t) = t^2 \sin(t), \quad y(t) = t^2 \cos(t)
\]

### Step 1: Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

\[
\frac{dx}{dt} = \frac{d}{dt}(t^2 \sin(t)) = 2t \sin(t) + t^2 \cos(t)
\]

\[
\frac{dy}{dt} = \frac{d}{dt}(t^2 \cos(t)) = 2t \cos(t) - t^2 \sin(t)
\]

### Step 2: Substitute into the Arc Length Formula

\[
S = \int_{0}^{4} \sqrt{\left(2t \sin(t) + t^2 \cos(t)\right)^2 + \left(2t \cos(t) - t^2 \sin(t)\right)^2} \, dt
\]

### Step 3: Simplify the Expression

First, expand each squared term:

\[
\left(2t \sin(t) + t^2 \cos(t)\right)^2 = 4t^2 \sin^2(t) + 4t^3 \sin(t) \cos(t) + t^4 \cos^2(t)
\]

\[
\left(2t \cos(t) - t^2 \sin(t)\right)^2 = 4t^2 \cos^2(t) - 4t^3 \sin(t) \cos(t) + t^4 \sin^2(t)
\]

Add these expressions:

\[
4t^2 \sin^2(t) + t^4 \cos^2(t) + 4t^2 \cos^2(t) + t^4 \sin^2(t) = 4t^2(\sin^2(t) + \cos^2(t)) + t^4(\sin^2(t) + \cos^2(t))
\]

Since \(\sin^2(t) + \cos^2(t) = 1\), we have:

\[
4t^2 + t^4
\]

Thus, the arc length formula becomes:

\[
S = \int_{0}^{4} \sqrt{4t^2 + t^4} \, dt
\]

Factor out \( t^2 \):

\[
S = \int_{0}^{4} t \sqrt{t^2 + 4} \, dt
\]

### Step 4: Evaluate the Integral

Let \( u = t^2 + 4 \), then \( du = 2t dt \), or \( t dt = \frac{1}{2} du \).

When \( t = 0 \), \( u = 4 \). When \( t = 4 \), \( u = 20 \).

\[
S = \frac{1}{2} \int_{4}^{20} \sqrt{u} \, du
\]

Integrate:

\[
S = \frac{1}{2} \int_{4}^{20} u^{1/2} du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{4}^{20}
\]

\[
S = \frac{1}{3} \left[ (20)^{3/2} - (4)^{3/2} \right]
\]

\[
S = \frac{1}{3} \left( 20\sqrt{20} - 8 \right)
\]

Therefore, the exact arc length is:

\[
S = \frac{20\sqrt{20} - 8}{3} \text{ units}
\]

Would you like me to explain any part of this process in more detail or provide further simplification?

### Related Questions:
1. What is the physical interpretation of arc length in parametric curves?
2. How does the arc length formula differ for polar coordinates?
3. What would the arc length be over a different interval, say \(0 \leq t \leq 2\)?
4. Can the integral for arc length always be expressed exactly, or are numerical methods sometimes necessary?
5. What are some examples of functions where calculating arc length is particularly difficult?

### Tip:
When working with parametric curves, always verify if the resulting integrals can be simplified before attempting to solve them directly; this often saves time and effort!

A curve is given by the following parametric equations. x(t)=t^2sin(t), y(t)=t^2cos(t). Find the arc length, S, of the curve over the interval 0 is less than or equal to t, is less than or equal to 4. Enter the exact value without decimals. S= _______ units.

Discover how to calculate the exact arc length of a parametric curve defined by x(t) = t²sin(t) and y(t) = t²cos(t) over the interval [0, 4]. This solution uses calculus, including differentiation and integration, to find the result. Perfect for students studying college-level calculus.

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