Find the exact length of the curve described by the parametric equations: $x = 7 + 3t^2, \quad y = 5 + 2t^3, \quad 0 \leq t \leq 5.$

Neetesh Kumar | January 3, 2025

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This is the solution to Math 1c
Assignment: 10.2 Question Number 13
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Step-by-step solution:

Step 1: Formula for the length of a parametric curve

The length of a curve described parametrically as $x = f(t)$ and $y = g(t)$ for $a \leq t \leq b$ is given by:

$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

Here, $a = 0$, $b = 5$, $x = 7 + 3t^2$, and $y = 5 + 2t^3$.

Step 2: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$

1. $x = 7 + 3t^2 \implies \frac{dx}{dt} = 6t$

2. $y = 5 + 2t^3 \implies \frac{dy}{dt} = 6t^2$

Step 3: Substitute into the formula

Substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the length formula:

$L = \int_0^5 \sqrt{(6t)^2 + (6t^2)^2} \, dt$

Simplify the terms inside the square root:

$L = \int_0^5 \sqrt{36t^2 + 36t^4} \, dt$

Factor out $36t^2$:

$L = \int_0^5 \sqrt{36t^2(1 + t^2)} \, dt$

Simplify further:

$L = \int_0^5 6t\sqrt{1 + t^2} \, dt$

Step 4: Use substitution to solve the integral

Let $u = 1 + t^2 \implies du = 2t \, dt$.

When $t = 0$, $u = 1$; and when $t = 5$, $u = 26$.

Rewriting the integral:

$L = \int_1^{26} 6 \cdot \frac{\sqrt{u}}{2} \, du$

Simplify:

$L = 3 \int_1^{26} \sqrt{u} \, du$#

Evaluate the integral:

$\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3}u^{3/2}$

Apply the limits:

$L = 3 \left[\frac{2}{3}u^{3/2}\right]_1^{26}$

Simplify:

$L = 2 \left[u^{3/2}\right]_1^{26}$

Substitute the limits:

$L = 2 \left[26^{3/2} - 1^{3/2}\right]$

Since $1^{3/2} = 1$, we have:

$L = 2 \left[26^{3/2} - 1\right] \approx 263.149014707$

Final Answer:

The exact length of the curve is:

$L = \boxed{2 \left[26^{3/2} - 1\right] \approx 263.149014707}$

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