onsider the curve $$ f(y)=\frac{1}{20} y^{4}+\frac{5}{8 y^{2}} $$ for $$ 1 \leq y \leq 3 $$. Find the arc length of the curve.

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UpStudy AI · Accepted Answer

To find the arc length of the curve $$ f(y) = \frac{1}{20} y^4 + \frac{5}{8 y^2} $$ for $$ 1 \leq y \leq 3 $$, we use the formula for arc length when the function is expressed as $$ x = f(y) $$:

$$
\text{Arc Length} = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
$$

### Step 1: Compute the Derivative $$\frac{dx}{dy}$$

Given:
$$
f(y) = \frac{1}{20} y^4 + \frac{5}{8 y^2}
$$

Differentiate with respect to $$ y $$:
$$
\frac{dx}{dy} = f'(y) = \frac{4}{20} y^3 - \frac{5}{4} y^{-3} = \frac{1}{5} y^3 - \frac{5}{4 y^3}
$$

### Step 2: Compute $$1 + \left(\frac{dx}{dy}\right)^2$$

$$
\left(\frac{dx}{dy}\right)^2 = \left(\frac{1}{5} y^3 - \frac{5}{4 y^3}\right)^2 = \frac{1}{25} y^6 - \frac{1}{2} + \frac{25}{16 y^6}
$$

Thus:
$$
1 + \left(\frac{dx}{dy}\right)^2 = \frac{1}{25} y^6 + \frac{1}{2} + \frac{25}{16 y^6}
$$

Notice that this expression is a perfect square:
$$
\frac{1}{25} y^6 + \frac{1}{2} + \frac{25}{16 y^6} = \left(\frac{1}{5} y^3 + \frac{5}{4 y^3}\right)^2
$$

So,
$$
\sqrt{1 + \left(\frac{dx}{dy}\right)^2} = \frac{1}{5} y^3 + \frac{5}{4 y^3}
$$

### Step 3: Integrate to Find the Arc Length

$$
\text{Arc Length} = \int_{1}^{3} \left(\frac{1}{5} y^3 + \frac{5}{4 y^3}\right) dy
$$

Integrate term by term:
$$
\int \frac{1}{5} y^3 \, dy = \frac{1}{20} y^4
$$
$$
\int \frac{5}{4 y^3} \, dy = \frac{5}{4} \left(-\frac{1}{2 y^2}\right) = -\frac{5}{8 y^2}
$$

So the definite integral is:
$$
\left[\frac{1}{20} y^4 - \frac{5}{8 y^2}\right]_{1}^{3}
$$

Evaluate from $$ y = 1 $$ to $$ y = 3 $$:
$$
\left(\frac{81}{20} - \frac{5}{72}\right) - \left(\frac{1}{20} - \frac{5}{8}\right) = \frac{41}{9}
$$

### Final Answer

The arc length of the curve from $$ y = 1 $$ to $$ y = 3 $$ is:

$$
\boxed{\dfrac{41}{9}}
$$
The arc length of the curve is $$ \frac{41}{9} $$.

onsider the curve for . Find the arc length of the curve.

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