Respuesta :
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: [tex]36 y^{2} =( x^{2} -4)^3 [/tex]
We divide by 36 and take the root of both sides to obtain: [tex]y = \sqrt{ \frac{( x^{2} -4)^3}{36} } [/tex]
Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: [tex]y = \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}[/tex]
Let's leave that for the moment and look at the formula for arc length. The formula is [tex]L= \int\limits^c_d {ds}[/tex] 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: [tex]ds= \sqrt{1+( \frac{dy}{dx})^2 } dx [/tex]
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 [tex]x^2-4[/tex]). More formally, we can let [tex]u=x^{2} -4[/tex] and then consider the derivative of [tex]u^{3/2}du[/tex]. Either way, we obtain,
[tex] \frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2} [/tex]
Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
[tex]( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4} [/tex]
This means that in our case:
[tex]ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx[/tex]
[tex]ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx[/tex]
[tex]ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx[/tex]
[tex]ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx[/tex]
[tex]ds= \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx [/tex]
Recall, the formula for arc length: [tex]L= \int\limits^c_d {ds}[/tex]
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:
[tex]L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x[/tex] 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, [tex][(\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}[/tex]
We are given: [tex]36 y^{2} =( x^{2} -4)^3 [/tex]
We divide by 36 and take the root of both sides to obtain: [tex]y = \sqrt{ \frac{( x^{2} -4)^3}{36} } [/tex]
Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: [tex]y = \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}[/tex]
Let's leave that for the moment and look at the formula for arc length. The formula is [tex]L= \int\limits^c_d {ds}[/tex] 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: [tex]ds= \sqrt{1+( \frac{dy}{dx})^2 } dx [/tex]
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 [tex]x^2-4[/tex]). More formally, we can let [tex]u=x^{2} -4[/tex] and then consider the derivative of [tex]u^{3/2}du[/tex]. Either way, we obtain,
[tex] \frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2} [/tex]
Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
[tex]( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4} [/tex]
This means that in our case:
[tex]ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx[/tex]
[tex]ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx[/tex]
[tex]ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx[/tex]
[tex]ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx[/tex]
[tex]ds= \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx [/tex]
Recall, the formula for arc length: [tex]L= \int\limits^c_d {ds}[/tex]
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:
[tex]L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x[/tex] 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, [tex][(\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}[/tex]
The exact value of curve is [tex]\dfrac{290}{3}[/tex].
[tex]36y^2 =(x^2-4)^3[/tex]
[tex]y^2=\dfrac{(x^2-4)^3}{36}[/tex]
[tex]y=\sqrt\dfrac{(x^2-4)^3}{36}[/tex]
[tex]y=\dfrac{(x^2-4)^{3/2}}{6}[/tex]-------eq [tex]1[/tex]
The formula of arc length ;
[tex]L=\int_{d}^{c}\sqrt{1+\dfrac{dy}{dx}^2} dx[/tex]-------eq [tex]2[/tex]
Find [tex]\dfrac{dy}{dx}[/tex]
Differentiate eq [tex]1[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{1}{6}\sqrt{(x^2-4)}(2x)[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{x}{2}\sqrt{(x^2-4)}[/tex]
Now square the above square equation,
[tex]\dfrac{dy}{dx}^2=\dfrac{x^2}{4}(x^2-4)[/tex]
[tex]\dfrac{dy}{dx}^2=\dfrac{x^4-4x^2}{4}[/tex]
so from arc length formula put [tex]\frac{\partial y}{\partial x}^2[/tex]in eq [tex]1[/tex]
[tex]L=\int_{d}^{c}\sqrt{1+\dfrac{x^4-4x^2}{4}} dx[/tex]
[tex]L=\int_{d}^{c}\sqrt{\dfrac{4+x^4-4x^2}{4}} dx[/tex]
[tex]L=\int_{d}^{c}{\dfrac{x^2-2}{2}} dx[/tex]
The limit of integration is given by 5 and 9
[tex]L=\int_{5}^{9}\dfrac{1}{2}{(x^2-1)}dx[/tex]
[tex]L=\int_{5}^{9}\dfrac{1}{2}(\dfrac{x^3}{3})-x[/tex]
Hence put values,
[tex]L=\dfrac{1}{2}(\dfrac{9^3}{3})-9- \dfrac{1}{2}(\dfrac{5^3}{3})-5[/tex]
[tex]L=(\dfrac{729}{6}-9 )(\dfrac{125}{6}-5 )[/tex]
[tex]L=\dfrac{290}{3}[/tex]
Learn more about integration here;
https://brainly.com/question/22008756?referrer=searchResults
