Question

Determine the length of y =(3x +2)^2 1<x<4

Answer 1

We have a equation of the form $y=f(x)$. To start, we'll find the derivative of the equation.

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(3x+2)^2$$

Using the chain rule, we can solve it to

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x+2)^2 = 2(3x+2)\cdot 3 = 18x+12$$

The length of a curve is given by

$$L=\int_a^b\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x$$

Substituting, we get

$$\begin{aligned} L&=\int_1^4\sqrt{1+\left(18x+12\right)^2}\mathrm{d}x\\ &=\int_1^4\sqrt{1+36\left(3x+2\right)^2}\mathrm{d}x\\ \end{aligned}$$

Substituting, $u=3x+2 \implies \frac{\mathrm{d}u}{\mathrm{d}x}=3 \implies \mathrm{d}x=\frac13\mathrm{d}u$

$$\begin{aligned} L &= \frac13\int\sqrt{1+36u^2}\mathrm{d}u\\ \end{aligned}$$

Substituting, $u=\frac{\tan v}6 \implies v=\arctan(6u) \implies \mathrm{d}u=\frac{\sec^2v}{6}\mathrm{d}v$

$$\begin{aligned} L &= \frac13\int\frac{\sec^2v\sqrt{1+\tan^2v}}{6}\mathrm{d}v\\ &= \frac1{18}\int\sec^3v\ \mathrm{d}v\\ \end{aligned}$$

This is a trivial integration solvable by a reduction formula and the standard formula for $\sec x$.

$$\int\sec^nx\ \mathrm{d}x = \frac{\sec^{n-2}x \tan x}{n-1} + \frac{n-2}{n-1}\int\sec^{n-2}x\ \mathrm{d}x$$

$$\begin{aligned} \therefore \int \sec^3v\ \mathrm{d}v &= \frac{\sec v \tan v}{2} + \frac{1}{2}\int\sec v\ \mathrm{d}v\\ &= \frac{\sec v \tan v}{2} + \frac{1}{2}\ln(\sec v + \tan v)\\ \end{aligned}$$

The final value is (ignoring the constant of integration)

$$\begin{aligned} L &= \frac1{18}\frac{\sec v \tan v + \ln(\tan v + \sec v)}{2}\\ &= \frac{\sqrt{1+36u^2}\cdot 6u + \ln(6u + \sqrt{1+36u^2})}{36}\\ &= \frac{\sqrt{1+36(3x+2)^2}\cdot 6(3x+2) + \ln(6(3x+2) + \sqrt{1+36(3x+2)^2})}{36}\\ \end{aligned}$$

Simplifying, and applying the limits, we get

$$\begin{aligned} L&=\left[ \frac{\ln(|\sqrt{324x^2+432x+145}+18x+12|)+(18x+12)\sqrt{324x^2+432x+145}}{36} \right]_1^4\\ &= \frac{\ln\left(\frac{84+\sqrt{7057}}{30+\sqrt{901}}\right)+ 6(14\sqrt{7057}-5\sqrt{901})}{36}\\ &\approx 171.0285971744902 \end{aligned}$$