A curve is given by the following parametric equations. X(T)=9tsin(t), y(t)=9tcos(t). Find the arc length, S, of the curve over the interval [0,1]. Enter the exact value with no decimals

To find the arc length \( S \) of a curve given by parametric equations \( x(t) \) and \( y(t) \), we use the following formula for parametric curves:

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

For the given parametric equations:

\[
x(t) = 9t\sin(t), \quad y(t) = 9t\cos(t)
\]

### Step 1: Differentiate \( x(t) \) and \( y(t) \)

\[
\frac{dx}{dt} = \frac{d}{dt}(9t\sin(t)) = 9(\sin(t) + t\cos(t))
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(9t\cos(t)) = 9(\cos(t) - t\sin(t))
\]

### Step 2: Compute the square of the derivatives

\[
\left(\frac{dx}{dt}\right)^2 = [9(\sin(t) + t\cos(t))]^2 = 81(\sin(t) + t\cos(t))^2
\]
\[
\left(\frac{dy}{dt}\right)^2 = [9(\cos(t) - t\sin(t))]^2 = 81(\cos(t) - t\sin(t))^2
\]

### Step 3: Simplify the sum

Now, we add these two expressions:

\[
\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 81[(\sin(t) + t\cos(t))^2 + (\cos(t) - t\sin(t))^2]
\]

Expand the terms inside:

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

Adding these two results gives:

\[
\sin^2(t) + \cos^2(t) + t^2(\sin^2(t) + \cos^2(t)) = 1 + t^2
\]

Thus, we have:

\[
\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 81(1 + t^2)
\]

### Step 4: Integrate to find the arc length

Now, we can find the arc length by integrating:

\[
S = \int_0^1 \sqrt{81(1 + t^2)} \, dt = 9\int_0^1 \sqrt{1 + t^2} \, dt
\]

This integral is a standard form, and its solution is:

\[
\int \sqrt{1 + t^2} \, dt = \frac{1}{2}t\sqrt{1+t^2} + \frac{1}{2}\ln(t + \sqrt{1+t^2})
\]

### Step 5: Evaluate the definite integral

Now, evaluate the integral from \( t = 0 \) to \( t = 1 \):

\[
S = 9 \left[ \frac{1}{2}(1\sqrt{2}) + \frac{1}{2}\ln(1 + \sqrt{2}) - \left(0 + \frac{1}{2}\ln(0 + 1)\right) \right]
\]

Simplifying:

\[
S = 9 \left( \frac{\sqrt{2}}{2} + \frac{1}{2}\ln(1 + \sqrt{2}) \right)
\]

Thus, the exact arc length is:

\[
S = \frac{9\sqrt{2}}{2} + \frac{9}{2}\ln(1 + \sqrt{2})
\]

### Final Answer:

The arc length is:

\[
S = \frac{9\sqrt{2}}{2} + \frac{9}{2}\ln(1 + \sqrt{2})
\]

Would you like more details or have any questions?

### Related Questions:

1. How does parametric differentiation apply in other contexts like velocity or acceleration?
2. What are the common applications of arc length in physics or engineering?
3. How would the integral change if the parameterization involved polar coordinates?
4. Can you find the arc length of a simpler parametric curve like \( x(t) = t^2 \), \( y(t) = t^3 \)?
5. What are the challenges in solving integrals involving square roots, and how can they be overcome?

**Tip:** Understanding parametric curves helps in analyzing complex motion, especially in multi-dimensional spaces.

A curve is given by the following parametric equations. X(T)=9tsin(t), y(t)=9tcos(t). Find the arc length, S, of the curve over the interval [0,1]. Enter the exact value with no decimals.

Discover how to find the exact arc length of the parametric curve given by X(T) = 9t*sin(t) and Y(T) = 9t*cos(t) over the interval [0, 1]. This problem involves calculus concepts like parametric differentiation and integral evaluation, suitable for advanced high school or college-level students.

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