Answer to Question #97139 in Calculus for Anand

Question

Find the length of the curve, 2y² = x³ from the vertex (0,0) to the point (4,4√2).

Answer 1

Solution:

We are going to find the length of the curve.

Given curve is

$$2 y^2 = x^3$$

We can write this for y;

$$y = ± \frac {1} {\sqrt 2} x^{\frac {3}{2}}$$

Differentiate with respect to x,

$$\frac {dy}{dx} = ± \frac {1} {\sqrt 2}(x^{\frac {3}{2}})^{\prime}$$

$$= ± \frac {1} {\sqrt 2} \times \frac {3} {2} \times x^{\frac {1}{2}} == ± \frac {3} {2\sqrt 2} \times x^{\frac {1}{2}}$$

We can find the length of the given curve $2 y^2 = x^3$ from (0, 0) to the point (4,4√2) by using a formula.

That is,

Length of the curve =

$$=\int_0^ 4 \sqrt {(1+ (\frac {dy}{dx})^2 } \space dx$$

$$= \int_0^ 4 \sqrt {(1+ (± \frac {3} {2\sqrt 2} \times x^{\frac {1}{2}})^2 } \space dx$$

$$= \int_0^ 4 \sqrt {(1+ ( \frac {9} {8} \times x) } \space dx$$

$$= \frac {1} {2\sqrt 2 } \int_0^4 \sqrt {8+ {9} x } \space dx$$

$$Let \space 8 +9x = t\\ 9 dx = dt \\ dx = \frac {1}{9} dt$$

$$Limits : x = 0 \space then \space t = 8 \\ x = 4 \space then \space t = 44$$

$$= \frac {1} {2\sqrt 2 } \int_8^{44} \sqrt t \space \frac {1}{9} \space dt$$

$$= \frac {1} {18\sqrt 2 } \int_8^{44} t^{\frac {1}{2}} \space \space dt$$

$$= \frac {1} {18\sqrt 2 } \space [\frac {t^{\frac {1}{2} + 1}} {\frac {1}{2} +1}]_8^{44}$$

$$= \frac {1} {18\sqrt 2 } \space \frac {2}{3} ((44)^{\frac {3}{2}} - (8)^\frac{3}{2})$$

$$= \frac {1} {27\sqrt 2 } (88\sqrt {11} - 16 \sqrt 2)$$

Answer:

$$Length \space of \space the \space curve= \frac {1} {27\sqrt 2 } (88\sqrt {11} - 16 \sqrt 2)$$