Answer:
Arc length = 168
Step-by-step explanation:
The arc length L of a curve y = f(x) over the interval [a, b] is given by the formula:
[tex]\displaystyle L=\int^b_a \sqrt{1+\left(\dfrac{\text{d}y}{\text{d}x}\right)^2}\; \text{d}x[/tex]
where a is the lower limit and b is the upper limit.
Given function:
[tex]y=\dfrac{1}{3}x^{\frac32}[/tex]
To find the length of the curve represented by the given function over the interval [0, 60], begin by differentiating y with respect to x:
[tex]\dfrac{dy}{dx}=\dfrac{3}{2} \cdot \dfrac{1}{3}x^{\frac32 - 1}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{1}{2}x^{\frac12}[/tex]
Now, substitute dy/dx into the arc length formula and integrate over the interval [0, 60]:
[tex]\displaystyle L=\int^{60}_0 \sqrt{1+\left(\dfrac{1}{2}x^{\frac12}\right)^2}\; dx[/tex]
[tex]\displaystyle L=\int^{60}_0 \sqrt{1+\dfrac{x}{4}}\; dx[/tex]
[tex]\displaystyle L=\int^{60}_0 \sqrt{\dfrac{4+x}{4}}\; dx[/tex]
[tex]\displaystyle L=\int^{60}_0 \dfrac{\sqrt{4+x}}{2}\; dx[/tex]
[tex]\displaystyle L=\int^{60}_0 \dfrac{1}{2}(4+x)^{\frac12}\; dx[/tex]
[tex]L=\left[\dfrac{1}{2} \cdot \dfrac{2}{3}(4+x)^{\frac32}\right]^{60}_0[/tex]
[tex]L=\left[\dfrac{1}{3}(4+x)^{\frac32}\right]^{60}_0[/tex]
[tex]L=\left(\dfrac{1}{3}(4+60)^{\frac32}\right)-\left(\dfrac{1}{3}(4+0)^{\frac32}\right)[/tex]
[tex]L=\dfrac{512}{3}-\dfrac{8}{3}[/tex]
[tex]L=\dfrac{504}{3}=168[/tex]
Therefore, the length of the curve of the given function on the interval [0, 60] is:
[tex]\Large\boxed{\boxed{168}}[/tex]
