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

over the interval

Ask by Malone Bowen. in the United States

Mar 29,2025

Question

Use your calculator to approximate (to the nearest thousandth) the length of the curve: $$ y=x^{\frac{2}{3}} $$ over the interval $$ x \in[1,8] $$ $$ [?] $$

UpStudy AI · Accepted Answer

We start with the general 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 given curve
$$
y=x^{\frac{2}{3}},
$$
we first compute the derivative:
$$
\frac{dy}{dx}=\frac{2}{3}x^{-\frac{1}{3}}.
$$
Squaring the derivative gives:
$$
\left(\frac{dy}{dx}\right)^2=\frac{4}{9}x^{-\frac{2}{3}}.
$$
Thus, the integrand is:
$$
\sqrt{1+\frac{4}{9}x^{-\frac{2}{3}}}.
$$
The arc length over the interval $$x\in[1,8]$$ is then:
$$
L=\int_{1}^{8}\sqrt{1+\frac{4}{9}x^{-\frac{2}{3}}}\,dx.
$$

To simplify the integral, we perform the substitution:
$$
u=x^{\frac{1}{3}} \quad\Longrightarrow\quad x=u^3,\quad dx=3u^2\,du.
$$
When $$x=1$$, then $$u=1$$; and when $$x=8$$, then $$u=2$$. Notice also that
$$
x^{-\frac{2}{3}}=u^{-2}.
$$
Substituting into the integrand:
$$
\sqrt{1+\frac{4}{9}u^{-2}}=\sqrt{\frac{9u^2+4}{9u^2}}=\frac{\sqrt{9u^2+4}}{3u}.
$$
Now, replacing $$dx$$ by $$3u^2\,du$$, the integral becomes:
$$
L=\int_{u=1}^{2} \frac{\sqrt{9u^2+4}}{3u}\cdot 3u^2\,du=
\int_{1}^{2} u\sqrt{9u^2+4}\,du.
$$

Next, we solve the integral:
$$
I = \int_{1}^{2} u\sqrt{9u^2+4}\,du.
$$
Make the substitution:
$$
w=9u^2+4 \quad\Longrightarrow\quad dw=18u\,du \quad\Longrightarrow\quad u\,du=\frac{dw}{18}.
$$
When $$u=1$$, $$w=9(1)^2+4=13$$; when $$u=2$$, $$w=9(4)+4=40$$.

The integral $$I$$ transforms to:
$$
I=\int_{w=13}^{40}\sqrt{w}\cdot\frac{dw}{18}=\frac{1}{18}\int_{13}^{40}w^{\frac{1}{2}}\,dw.
$$
Evaluate the integral:
$$
\int w^{\frac{1}{2}}\,dw=\frac{2}{3}w^{\frac{3}{2}}.
$$
Thus,
$$
I=\frac{1}{18}\cdot\frac{2}{3}\left[w^{\frac{3}{2}}\right]_{13}^{40}=\frac{1}{27}\left(40^{\frac{3}{2}}-13^{\frac{3}{2}}\right).
$$
This gives the arc length:
$$
L=\frac{1}{27}\left(40^{\frac{3}{2}}-13^{\frac{3}{2}}\right).
$$

Now we approximate the value numerically. Compute:
$$
40^{\frac{3}{2}}=(\sqrt{40})^3\quad \text{and} \quad 13^{\frac{3}{2}}=(\sqrt{13})^3.
$$
Approximately,
$$
\sqrt{40}\approx6.3246 \quad\Longrightarrow\quad 40^{\frac{3}{2}}\approx6.3246^3\approx252.982,
$$
$$
\sqrt{13}\approx3.6056 \quad\Longrightarrow\quad 13^{\frac{3}{2}}\approx3.6056^3\approx46.872.
$$
Subtracting and dividing by 27:
$$
L\approx\frac{252.982-46.872}{27}=\frac{206.110}{27}\approx7.630.
$$

Thus, the approximate length of the curve is:
$$
\boxed{7.630}.
$$
The approximate length of the curve $$ y = x^{\frac{2}{3}} $$ from $$ x = 1 $$ to $$ x = 8 $$ is **7.630**.

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