Use your calculator to approximate
(to the nearest thousandth) the
length of the curve:

over the interval

Ask by Bowers Marshall. in the United States

Mar 29,2025

Question

Use your calculator to approximate (to the nearest thousandth) the length of the curve: $$ y=\frac{1}{3}\left(x^{2}+2\right)^{\frac{3}{2}} $$ over the interval $$ x \in[1,3] $$ $$ [?] $$

UpStudy AI · Accepted Answer

We start with the formula for the arc length of a curve $$ y=f(x) $$ over $$[a,b]$$:

$$
L=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx
$$

For the curve

$$
y=\frac{1}{3}\left(x^2+2\right)^{\frac{3}{2}},
$$

we first find its derivative.

**Step 1. Compute the derivative $$ y'=\frac{dy}{dx} $$:**

We use the chain rule. Write:

$$
y=\frac{1}{3}\left(x^2+2\right)^{\frac{3}{2}}.
$$

Differentiate with respect to $$ x $$:

$$
y'=\frac{1}{3}\cdot\frac{3}{2}\left(x^2+2\right)^{\frac{1}{2}}\cdot\frac{d}{dx}(x^2+2).
$$

Since

$$
\frac{d}{dx}(x^2+2)=2x,
$$

we have

$$
y'=\frac{1}{3}\cdot\frac{3}{2}\cdot 2x\left(x^2+2\right)^{\frac{1}{2}}.
$$

Simplify the constants:

$$
\frac{1}{3}\cdot\frac{3}{2}\cdot 2x = x.
$$

Thus,

$$
y'=x\left(x^2+2\right)^{\frac{1}{2}}.
$$

**Step 2. Compute $$ \sqrt{1+\left(y'\right)^2} $$:**

We have

$$
\left(y'\right)^2=x^2\left(x^2+2\right).
$$

So,

$$
1+\left(y'\right)^2=1+x^2\left(x^2+2\right).
$$

Expanding $$ x^2\left(x^2+2\right) $$:

$$
x^2\left(x^2+2\right)=x^4+2x^2,
$$

which gives us:

$$
1+\left(y'\right)^2=x^4+2x^2+1.
$$

Notice that

$$
x^4+2x^2+1=(x^2+1)^2.
$$

Taking the square root:

$$
\sqrt{1+\left(y'\right)^2}=\sqrt{(x^2+1)^2}=x^2+1,
$$

since $$ x^2+1 $$ is always positive.

**Step 3. Set up the arc length integral:**

The length of the curve from $$ x=1 $$ to $$ x=3 $$ is

$$
L=\int_{1}^{3} \left(x^2+1\right)\, dx.
$$

**Step 4. Evaluate the integral:**

We integrate term-by-term:

$$
\int\left(x^2+1\right)dx=\int x^2\,dx+\int 1\,dx=\frac{x^3}{3}+x.
$$

Now, we evaluate from $$ x=1 $$ to $$ x=3 $$:

$$
L=\left[\frac{x^3}{3}+x\right]_{1}^{3}=\left(\frac{3^3}{3}+3\right)-\left(\frac{1^3}{3}+1\right).
$$

Compute the values:

$$
\frac{3^3}{3}+3=\frac{27}{3}+3=9+3=12,
$$
$$
\frac{1^3}{3}+1=\frac{1}{3}+1=\frac{1}{3}+\frac{3}{3}=\frac{4}{3}.
$$

Thus,

$$
L=12-\frac{4}{3}=\frac{36}{3}-\frac{4}{3}=\frac{32}{3}\approx 10.667.
$$

**Final Answer:**

$$
\boxed{10.667}
$$
The length of the curve is approximately 10.667.

Use your calculator to approximate
(to the nearest thousandth) the
length of the curve:

over the interval

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Use your calculator to approximate (to the nearest thousandth) the length of the curve: over the interval