Arc Length = $$\displaystyle \int_{x=a}^b \sqrt{1+\left[f'(x)\right]^2}\,dx$$

Arc Length = $$\displaystyle \int_{y=c}^d \sqrt{1+\left[g'(y)\right]^2}\,dy$$

#### Steps

- find $$f'(x)$$
- find $$\big[f'(x)\big]^2$$
- find $$1+\big[f'(x)\big]^2$$
- find $$\sqrt{1+\big[f'(x)\big]^2}$$
- $$\displaystyle \int_{x=a}^b{\sqrt{1+\left[f'(x)\right]^2}\,dx}$$
- hope to find an antiderivative
- or evaluate numerically with a computer program

#### Homework

- Sketch and find the length of the curve $$y=\frac{1}{3}(x^2+2)^\frac{3}{2}$$ from $$x=0$$ to $$x=3$$.
- $$y’=\frac{1}{3}\frac{3}{2} 2x (x^2+2)^\frac{1}{2}=x (x^2+2)^\frac{1}{2}$$
- $$(y’)^2 = x^2 (x^2+2)$$
- $$1+(y’)^2 =1+ x^2 (x^2+2)$$

$$= x^4 + 2x^2 +1 = (x^2+1)^2$$ - $$\sqrt{1+(y’)^2} = x^2+1$$
- $$\int_0^3 \left(x^2+1\right) \,dx=\bigg[\frac{x^3}{3}+x\bigg]_0^3=12$$

- Sketch and find the length of the curve $$y=x^\frac{3}{2}$$ from $$x=0$$ to $$x=4$$.
- Sketch and find the length of the curve $$x=\frac{y^3}{3}+\frac{1}{4y}$$ from $$y=1$$ to $$y=3$$.
- Sketch and find the length of the curve $$x=\frac{y^4}{4}+\frac{1}{8y^2}$$ from $$y=1$$ to $$y=2$$.
- Sketch and find the length of the curve $$x=\int_0^x \sqrt{\sec^4(t)-1}\,dt$$, $$-\frac{\pi}{4}\le x \le\frac{\pi}{4}$$
- $$x’=\sqrt{\sec^4(x)-1}$$
- $$(x’)^2=\sec^4(x)-1$$
- $$1+(x’)^2 =\sec^4(x)$$
- $$\sqrt{1+(x’)^2} =\sec^2(x)$$
- $$\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{1+(x’)^2}\,dx =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2(x)\,dx=\bigg[\tan(x)\bigg]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=2$$

- $$y=x^2$$ , $$-1 \le x \le 2$$

Sketch the graph and set up the integral for the length of the curve. Use a calculator or program to approximate the curve’s length. - 13
- 14
- 15
- 21
- Find the arc length of an astroid. $$\displaystyle x^\frac{2}{3}+y^\frac{2}{3}=a^\frac{2}{3}$$ is a curve called an astroid, star-shaped. Solving for $$y$$ we get $$\displaystyle y=\left(a^\frac{2}{3}-x^\frac{2}{3}\right)^\frac{3}{2}$$. We will start by finding the arclength of half of the section in the first quadrant $$\frac{a\sqrt{2}}{4}$$.

In the graph to the right a = 3. You can see there that Geogebra had a difficult time drawing the cusps on each axis.

Solution - 23

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