$\text { Find the arc length of the graph of the function } y = \frac { 3 } { 2 } x ^ { \frac { 2 } { 3 } } + 10 \text { over the interval }$ [1,64].
A) $17 \sqrt [ 3 ] { 17 } - 2 \sqrt [ 3 ] { 2 }$
B) $19 \sqrt { 19 } - 2 \sqrt { 2 }$
C) $17 \sqrt { 17 } - 2 \sqrt { 2 }$
D) $19 \sqrt [ 3 ] { 19 } - 10 \sqrt [ 3 ] { 10 }$
E) $17 \sqrt { 17 } - 10 \sqrt { 10 }$

Question

$\text { Find the arc length of the graph of the function } y = \frac { 3 } { 2 } x ^ { \frac { 2 } { 3 } } + 10 \text { over the interval }$ [1,64]. 
A) $17 \sqrt [ 3 ] { 17 } - 2 \sqrt [ 3 ] { 2 }$
B) $19 \sqrt { 19 } - 2 \sqrt { 2 }$
C) $17 \sqrt { 17 } - 2 \sqrt { 2 }$
D) $19 \sqrt [ 3 ] { 19 } - 10 \sqrt [ 3 ] { 10 }$
E) $17 \sqrt { 17 } - 10 \sqrt { 10 }$

Quizplus · Accepted Answer

The correct answer is C because the arc length of the graph of the function is given by the integral of the square root of 1 plus the square of the derivative of the function. The derivative of the function is y' = x^(1/3), so the arc length is given by the integral of the square root of 1 + x^(2/3) from x = 1 to x = 64. This integral can be evaluated using the substitution u = 1 + x^(2/3), which gives du/dx = (2/3)x^(-1/3) and dx = (3/2)u^(-1/2)du. Substituting into the integral, we get:$$\int_1^{64} \sqrt{1 + x^{2/3}} dx = \int_{1 + 1^{2/3}}^{1 + 64^{2/3}} \sqrt{u} \frac{3}{2}u^{-1/2} du = \frac{3}{2} \int_{1 + 1^{2/3}}^{1 + 64^{2/3}} u^{1/2} du$$$$= \frac{3}{2} \left[\frac{2}{3}u^{3/2}\right]_{1 + 1^{2/3}}^{1 + 64^{2/3}} = \frac{1}{3} \left[(1 + 64^{2/3})^{3/2} - (1 + 1^{2/3})^{3/2}\right]$$$$= \frac{1}{3} \left[(1 + 64^{2/3})^{3/2} - 2^{3/2}\right] = \frac{1}{3} \left[17^{3/2} - 2^{3/2}\right] = \frac{1}{3} \left[17\sqrt{17} - 2\sqrt{2}\right]$$$$= 17\sqrt{17} - 2\sqrt{2}$$