Length of Curve $y^2=4(x+4)^3$: 13.5429

In summary, the conversation includes a problem involving finding the derivative and integral of a given equation, with the final answer being 13.5429. The person initially doubted their answer but it was confirmed to be correct. Another person suggests a way to get an exact answer for the integral. The conversation ends with the person thanking the other for their help.
  • #1
ineedhelpnow
651
0
$y^2=4(x+4)^3$
$0 \le x \le 2$

$y=2(x+4)^{3/2}$

$y'=3(x+4)^{1/2}$

$\int_{0}^{2} \ \sqrt{1+9(x+4)},dx = 13.5429$

is that right?
 
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  • #2
wait, no one respond yet. i wrote the question down wrong. i have to back and redo it.

actually that is the right question. did i do it right?
 
Last edited:
  • #3
ineedhelpnow said:
$y^2=4(x+4)^3$
$0 \le x \le 2$

$y=2(x+4)^{3/2}$

$y'=3(x+4)^{1/2}$

$\int_{0}^{2} \ \sqrt{1+9(x+4)}\,dx = 13.5429$

is that right?

What you wrote is all correct (assuming $y\ge 0$).

What has your final integral got to do with what came before it?
 
  • #4
$\int_{a}^{b} \ \sqrt{1+[f'(x)]^2},dx$

the steps i showed was solving for my y' and then i squared it and plugged it in.
 
  • #5
ineedhelpnow said:
$y^2=4(x+4)^3$
$0 \le x \le 2$

$y=2(x+4)^{3/2}$

$y'=3(x+4)^{1/2}$

$\int_{0}^{2} \ \sqrt{1+9(x+4)},dx = 13.5429$

is that right?

Your setup of the integral is correct. You can probably get an exact answer though.

$\displaystyle \begin{align*} \int_0^2{ \sqrt{ 1 + 9 \left( x + 4 \right) } \,\mathrm{d}x } &= \int_0^2{ \sqrt{ 1 + 9x + 36 }\,\mathrm{d}x } \\ &= \int_0^2{ \sqrt{ 9x + 37 } \, \mathrm{d}x } \\ &= \frac{1}{9} \int_0^2{ 9\,\sqrt{ 9x + 37 } \, \mathrm{d}x } \end{align*}$

and now make the substitution $\displaystyle \begin{align*} u = 9x + 37 \implies \mathrm{d}u = 9\,\mathrm{d}x \end{align*}$ and noting that $\displaystyle \begin{align*} u(0) = 37 \end{align*}$ and $\displaystyle \begin{align*} u(2) = 55 \end{align*}$, the integral becomes

$\displaystyle \begin{align*} \frac{1}{9} \int_0^2{ 9\,\sqrt{ 9x + 37} \, \mathrm{d}x } &= \frac{1}{9} \int_{37}^{55}{ \sqrt{u}\,\mathrm{d}u } \\ &= \frac{1}{9} \int_{37}^{55}{ u^{\frac{1}{2}}\,\mathrm{d}u } \\ &= \frac{1}{9} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right] _{37}^{55} \\ &= \frac{1}{9} \left[ \frac{2}{3} u\,\sqrt{u} \right] _{37}^{55} \\ &= \frac{2}{27} \left( 55\,\sqrt{55} - 37\,\sqrt{37} \right) \end{align*}$
 
  • #6
thats what i got but i just posted the decimal form because it was easier. thanks for working it out though :)
 

FAQ: Length of Curve $y^2=4(x+4)^3$: 13.5429

What is the equation for this curve?

The equation for this curve is y^2=4(x+4)^3: 13.5429. This is a cubic curve in the form of y^2=4ax^3 where a is the coefficient of the curve.

What is the length of this curve?

The length of the curve y^2=4(x+4)^3: 13.5429 can be calculated using the arc length formula for a parametric curve. This formula is given by L = ∫√(1+(dy/dx)^2) dx. By solving this integral, we can find the length of the curve.

How do you find the coordinates of the curve?

The coordinates of the curve can be found by substituting different values of x into the equation y^2=4(x+4)^3: 13.5429. This will give us corresponding values of y, which we can plot on a graph to get the coordinates of the curve.

What is the shape of this curve?

The shape of this curve is a cubic curve. It is a parabola-like shape with a slight twist, and it is symmetric about the y-axis. The curve also has a point of inflection at the origin, where it changes from concave up to concave down.

How is the length of this curve related to its equation?

The length of this curve is directly related to its equation. As the equation changes, the shape and length of the curve will also change. In this case, the coefficient a in the equation affects the steepness of the curve, which in turn affects the length. A larger value of a will result in a steeper curve and a longer length, while a smaller value of a will result in a flatter curve and a shorter length.

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