Solutions to Problem Set 32

Math 211-03

11-27-2017

[Arc length]

1. Find the length of the curve

$$x = 2 y^{3/2}, \quad 0 \le y \le 1.$$

$$\der x y = 3 y^{1/2}, \quad 1 + \left(\der x y\right)^2 = 1 + 9 y.$$

The length is

$$\int_0^1 \sqrt{1 + 9 y}\,dy = \left[\dfrac{2}{27} (1 + 9 y)^{3/2}\right]_0^1 = \dfrac{2}{27} (10^{3/2} - 1) = 2.26835 \ldots .\quad\halmos$$


2. Find the length of the curve

$$y = \dfrac{1}{5} x^5 + \dfrac{1}{12} \dfrac{1}{x^3}, \quad 1 \le x \le 2.$$

$$1 + \left(\der y x\right)^2 = 1 + \left(x^4 - \dfrac{1}{4} \dfrac{1}{x^4}\right)^2 = 1 + x^8 - \dfrac{1}{2} + \dfrac{1}{16} \dfrac{1}{x^8} = x^8 + \dfrac{1}{2} + \dfrac{1}{16} \dfrac{1}{x^8} = \left(x^4 + \dfrac{1}{4} \dfrac{1}{x^4}\right)^2.$$

The length is

$$\int_1^2 \left(x^4 + \dfrac{1}{4} \dfrac{1}{x^4}\right)\,dx = \left[\dfrac{1}{5} x^5 - \dfrac{1}{12} \dfrac{1}{x^3}\right]_1^2 = \dfrac{3011}{480} = 6.27291 \ldots.\quad\halmos$$


3. Find the length of the curve

$$x = \dfrac{2}{3} y^{3/2} + 1, \quad 0 \le y \le 1.$$

$$1 + \left(\der x y\right)^2 = 1 + (y^{1/2})^2 = 1 + y.$$

The length is

$$\int_0^1 \sqrt{1 + y}\,dy = \left[\dfrac{2}{3} (1 + y)^{3/2}\right]_0^1 = \dfrac{2}{3} (2^{3/2} - 1) = 1.21895 \ldots.\quad\halmos$$


4. Find the length of the curve

$$x = e^{3 t} + e^{-3 t} + 2, \quad y = 6 t + 1, \quad 0 \le t \le 1.$$

$$\left(\der x t\right)^2 + \left(\der y t\right)^2 = (3 e^{3 t} - 3 e^{-3 t})^2 + 6^2 = 9 e^{6 t} - 18 + 9 e^{-6 t} + 36 =$$

$$9 e^{6 t} + 18 + 9 e^{-6 t} = (3 e^{3 t} + 3 e^{-3 t})^2.$$

The length is

$$\int_0^1 \left(3 e^{3 t} + 3 e^{-3 t}\right)\,dt = \left[e^{3 t} - e^{-3 t}\right]_0^1 = e^3 - e^{-3} = 20.03574 \ldots.\quad\halmos$$


5. Find the length of the curve

$$x = -\dfrac{1}{2} e^{-2 t} - 16 t + 3, \quad y = 8 e^{-t} + 1, \quad 0 \le t \le 1.$$

$$\left(\der x t\right)^2 + \left(\der y t\right)^2 = (e^{-2 t} - 16)^2 + (-8 e^{-t})^2 = e^{-4 t} - 32 e^{-2 t} + 256 + 64 e^{-2 t} =$$

$$e^{-4 t} + 32 e^{-2 t} + 256 = (e^{-2 t} + 16)^2.$$

The length is

$$\int_0^1 \left(e^{-2 t} + 16\right)\,dt = \left[-\dfrac{1}{2} e^{-2 t} + 16 t\right]_0^1 = \dfrac{33}{2} - \dfrac{1}{2} e^{-2} = 16.43233 \ldots.\quad\halmos$$


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