Question:

Let $C$ be the curve defined by the equation $y = \frac{1}{2}x^3 - x^2 - x$.

a) Find the length of the curve $C$ between $x=0$ and $x=2$. Show all work.

b) Find the length of the curve $C$ between $x=0$ and $x=4$. Show all work.

Answer:

a) To find the length of the curve $C$ between $x=0$ and $x=2$, we can use the arc length formula for a function $f(x)$. The formula is given by:

where $f'(x)$ represents the derivative of $f(x)$.

Given that $y = \frac{1}{2}x^3 - x^2 - x$, let's find the derivative $y'$:

Now, we can calculate the integrand $\sqrt{1 + (y')^2}$:

Let's integrate $\sqrt{1 + (y')^2}$ over the interval $x=0$ to $x=2$:

This integral is difficult to evaluate directly. We can simplify it using trigonometric substitution. Let:

Taking the derivative of both sides, we get:

Solving for $\frac{dt}{dx}$, we get:

Rearranging, we have:

Substituting the values into the integral, we get:

where $t_1$ and $t_2$ are the corresponding limits for $x=0$ and $x=2$ respectively.

To find $t_1$ and $t_2$, we solve for $x$ in terms of $t$:

This is a quadratic equation that can be solved using the quadratic formula:

Setting $x=0$ and $x=2$ gives us:

$t_1 = \text{arcsinh}\left(\frac{\sqrt{6}}{2}\right)$ and $t_2 = \text{arcsinh}\left(\frac{2\sqrt{6}}{3}\right)$

After substituting the values, we get:

Unfortunately, this integral cannot be evaluated exactly. We need to either approximate using numerical methods or use a calculator or computer software to find the answer.

b) Similarly, to find the length of the curve $C$ between $x=0$ and $x=4$, we use the same arc length formula:

Using the same substitutions as before, we find:

$t_1 = \text{arcsinh}\left(\frac{\sqrt{10}}{2}\right)$ and $t_2 = \text{arcsinh}\left(\frac{4\sqrt{10}}{3}\right)$

Substituting the values into the formula, we have:

Again, this integral cannot be evaluated exactly, so numerical methods or technology are needed to find the answer.