# If the curve of y = f(x) has the property that y' = \sqrt{x^3 - 1} , finid the length of the...

## Question:

If the curve of y = f(x) has the property that {eq}y' = \sqrt{x^3 - 1} {/eq}, finid the length of the curve from x = 4 to x = 9.

## Arc Length of a Curve:

The arc length {eq}L {/eq} of a curve {eq}y=f(x) {/eq} in the interval {eq}[c,d] {/eq} is given by :

{eq}L=\int_{c}^{d}\sqrt{1+\left ( y' \right )^{2}} \ dx {/eq}

In the given problem, the derivative of the function is given to us

so, we directly use the above formula to find the arc length.

## Answer and Explanation:

Given that:

{eq}y' = \sqrt{ x^{3} - 1} {/eq}

where {eq}4 \leq x \leq 9 {/eq}

Therefore,

{eq}\\\\\begin{align*} \sqrt{1+(y')^{2}} & =\sqrt{1+\left(\sqrt{x^3 - 1} \right)^{2}} \\\\&= \sqrt{1+x^3 - 1} \\\\&=\sqrt{x^3} \\\\&=x^{3/2} ---(1) \\\\\end{align*} {/eq}

Thus, the required arc length of the curve is:

{eq}\\\\\begin{align*} \int_{4}^{9} ds & = \int_{4}^{9} \sqrt{1+(y')^{2}} \ dx \\\\& = \int_{4}^{9} x^{3/2} \ dx \ \ \ \texttt{ (Using }(1)) \\\\& = \left [ \frac{2x^{5/2}}{5} \right ]_{4}^{9} \\\\& = \frac{2(9)^{5/2}}{5}-\frac{2(4)^{5/2}}{5} \\\\& = \frac{2}{5}\left [(9)^{5/2}-(4)^{5/2} \right ] \\\\& = \frac{2}{5}\left [(3^{2})^{5/2}-(2^{2})^{5/2} \right ] \\\\& = \frac{2}{5}\left [(3)^{5}-(2)^{5} \right ] \\\\& = \frac{2}{5}\left [ 243-32 \right ] \\\\& = \frac{2}{5}\left [ 211 \right ] \\\\& = \frac{422}{5} \texttt{ units } \\\\\end{align*} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question