How do you find the length of the curve $y=(2x+1{)}^{\frac{3}{2}},0\le x\le 2$?

DIAMMIBENVERMk1
2022-07-04
Answered

How do you find the length of the curve $y=(2x+1{)}^{\frac{3}{2}},0\le x\le 2$?

You can still ask an expert for help

Ordettyreomqu

Answered 2022-07-05
Author has **22** answers

The length of a curve between a and b values for x is given by

$L={\int}_{a}^{b}\sqrt{1+{y}^{\prime 2}}dx$

${y}^{\prime 2}=18x+9$

$L={\int}_{0}^{2}\sqrt{18x+10}dx=\frac{1}{27}(18x+10{)}^{\frac{3}{2}}{]}_{0}^{2}=10.38$

$L={\int}_{a}^{b}\sqrt{1+{y}^{\prime 2}}dx$

${y}^{\prime 2}=18x+9$

$L={\int}_{0}^{2}\sqrt{18x+10}dx=\frac{1}{27}(18x+10{)}^{\frac{3}{2}}{]}_{0}^{2}=10.38$

pablos28spainzd

Answered 2022-07-06
Author has **4** answers

The Arc Length of a curve y=f(x) from x=a to x=b is given by:

$L={\int}_{a}^{b}\sqrt{1+(\frac{dy}{dx}{)}^{2}}dx$

So, for the given curve $y=(2x+1{)}^{\frac{3}{2}}$ for $x\in [0,2]$ we form the derivative using the power rule for differentiation in conjunction with the chain rule:

$\frac{dy}{dx}=(\frac{3}{2})(2x+1{)}^{\frac{3}{2}-1}(2)$

$=3(2x+1{)}^{\frac{1}{2}}$

So then, the arc length is:

$L={\int}_{0}^{2}\sqrt{1+(3(2x+1{)}^{\frac{1}{2}}{)}^{2}}dx$

$={\int}_{0}^{2}\sqrt{1+9(2x+1)}dx$

$={\int}_{0}^{2}\sqrt{1+18x+9}dx$

$={\int}_{0}^{2}\sqrt{18x+10}dx$

And using the power rule for integration, we can integrate to get:

$L=[\frac{(18x+10{)}^{\frac{3}{2}}}{\frac{3}{2}}\ast \frac{1}{18}{]}_{0}^{2}$

$=[\frac{1}{27}(18x+10{)}^{\frac{3}{2}}{]}_{0}^{2}$

$=\frac{1}{27}[(36+10{)}^{\frac{3}{2}}-{10}^{\frac{3}{2}}]$

$=\frac{1}{27}[46\sqrt{46}-10\sqrt{10}]$

$\approx 10.384(3dp)$

$L={\int}_{a}^{b}\sqrt{1+(\frac{dy}{dx}{)}^{2}}dx$

So, for the given curve $y=(2x+1{)}^{\frac{3}{2}}$ for $x\in [0,2]$ we form the derivative using the power rule for differentiation in conjunction with the chain rule:

$\frac{dy}{dx}=(\frac{3}{2})(2x+1{)}^{\frac{3}{2}-1}(2)$

$=3(2x+1{)}^{\frac{1}{2}}$

So then, the arc length is:

$L={\int}_{0}^{2}\sqrt{1+(3(2x+1{)}^{\frac{1}{2}}{)}^{2}}dx$

$={\int}_{0}^{2}\sqrt{1+9(2x+1)}dx$

$={\int}_{0}^{2}\sqrt{1+18x+9}dx$

$={\int}_{0}^{2}\sqrt{18x+10}dx$

And using the power rule for integration, we can integrate to get:

$L=[\frac{(18x+10{)}^{\frac{3}{2}}}{\frac{3}{2}}\ast \frac{1}{18}{]}_{0}^{2}$

$=[\frac{1}{27}(18x+10{)}^{\frac{3}{2}}{]}_{0}^{2}$

$=\frac{1}{27}[(36+10{)}^{\frac{3}{2}}-{10}^{\frac{3}{2}}]$

$=\frac{1}{27}[46\sqrt{46}-10\sqrt{10}]$

$\approx 10.384(3dp)$

asked 2021-12-20

Evaluate the following integrals. Include absolute values only when needed.

$\int \frac{{\mathrm{ln}}^{2}x+2\mathrm{ln}x-1}{x}dx$

asked 2021-12-13

Use partial fractions to evaluate the definite integral.

${\int}_{0}^{1}\frac{{x}^{2}-x}{{x}^{2}+x+1}dx$

asked 2022-04-16

What is a solution to the differential equation $\frac{dy}{dx}=\frac{{e}^{x}}{y}$ with y(0)=1?

asked 2022-06-22

How do you find the point c in the interval $0\le x\le 2$ such that f(c) is equation to the average value of $f(x)=\sqrt{2x}$?

asked 2022-04-11

How do you solve for xy'-y=3xy given y(1)=0?

asked 2022-05-30

How do you find the arc length of the curve $y=\mathrm{ln}\mathrm{cos}x$ over the interval $[0,\pi /3]$?

asked 2021-12-31

Evaluate the integrals.

$\int {x}^{3}{e}^{{x}^{4}}dx$