Mathematics, 27.07.2019 13:00 tordiacasey

Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0

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We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: $36 y^{2} =( x^{2} -4)^3$
We divide by 36 and take the root of both sides to obtain: $y = \sqrt{ \frac{( x^{2} -4)^3}{36} }$

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: $y = \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}$

Let's leave that for the moment and look at the formula for arc length. The formula is $L= \int\limits^c_d {ds}$ where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: $ds= \sqrt{1+( \frac{dy}{dx})^2 } dx$

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4

). More formally, we can let $u=x^{2} -4$ and then consider the derivative of $u^{3/2}du$ . Either way, we obtain,

$\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}$

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
$( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}$

This means that in our case:
$ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx$
$ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx$
$ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx$
$ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx$
$ds= \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx$

Recall, the formula for arc length: $L= \int\limits^c_d {ds}$
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

$L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x$ evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, $[(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}$

Answer from: ssiy

The exact value of curve is $\dfrac{290}{3}$ .

36y^2 =(x^2-4)^3

$y^2=\dfrac{(x^2-4)^3}{36}$

$y=\sqrt\dfrac{(x^2-4)^3}{36}$

$y=\dfrac{(x^2-4)^{3/2}}{6}$ -------eq

The formula of arc length ;

$L=\int_{d}^{c}\sqrt{1+\dfrac{dy}{dx}^2} dx$ -------eq

Find $\dfrac{dy}{dx}$

Differentiate eq

$\dfrac{dy}{dx}=\dfrac{1}{6}\sqrt{(x^2-4)}(2x)$

$\dfrac{dy}{dx}=\dfrac{x}{2}\sqrt{(x^2-4)}$

Now square the above square equation,

$\dfrac{dy}{dx}^2=\dfrac{x^2}{4}(x^2-4)$

$\dfrac{dy}{dx}^2=\dfrac{x^4-4x^2}{4}$

so from arc length formula put $\frac{\partial y}{\partial x}^2$ in eq

$L=\int_{d}^{c}\sqrt{1+\dfrac{x^4-4x^2}{4}} dx$

$L=\int_{d}^{c}\sqrt{\dfrac{4+x^4-4x^2}{4}} dx$

$L=\int_{d}^{c}{\dfrac{x^2-2}{2}} dx$

The limit of integration is given by 5 and 9

$L=\int_{5}^{9}\dfrac{1}{2}{(x^2-1)}dx$

$L=\int_{5}^{9}\dfrac{1}{2}(\dfrac{x^3}{3})-x$

Hence put values,

$L=\dfrac{1}{2}(\dfrac{9^3}{3})-9- \dfrac{1}{2}(\dfrac{5^3}{3})-5$

$L=(\dfrac{729}{6}-9 )(\dfrac{125}{6}-5 )$

$L=\dfrac{290}{3}$

Learn more about integration here;

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Answer from: Quest

step-by-step explanation:

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Answer from: Quest

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