Answer:
[tex]f(x) = \frac{6}{x^2} -8[/tex] or [tex]f(x) = -\frac{6}{x^2} + 4[/tex]
Step-by-step explanation:
Given
[tex](x,y) = (1,-2)[/tex] --- Point
[tex]\int\limits^6_2 {(1 + 144x^{-6})} \, dx[/tex]
The arc length of a function on interval [a,b]: [tex]\int\limits^b_a {(1 + f'(x^2))} \, dx[/tex]
By comparison:
[tex]f'(x)^2 = 144x^{-6}[/tex]
[tex]f'(x)^2 = \frac{144}{x^6}[/tex]
Take square root of both sides
[tex]f'(x) =\± \sqrt{\frac{144}{x^6}}[/tex]
[tex]f'(x) = \±\frac{12}{x^3}[/tex]
Split:
[tex]f'(x) = \frac{12}{x^3}[/tex] or [tex]f'(x) = -\frac{12}{x^3}[/tex]
To solve fo f(x), we make use of:
[tex]f(x) = \int {f'(x) } \, dx[/tex]
For: [tex]f'(x) = \frac{12}{x^3}[/tex]
[tex]f(x) = \int {\frac{12}{x^3} } \, dx[/tex]
Integrate:
[tex]f(x) = \frac{12}{2x^2} + c[/tex]
[tex]f(x) = \frac{6}{x^2} + c[/tex]
We understand that it passes through [tex](x,y) = (1,-2)[/tex].
So, we have:
[tex]-2 = \frac{6}{1^2} + c[/tex]
[tex]-2 = \frac{6}{1} + c[/tex]
[tex]-2 = 6 + c[/tex]
Make c the subject
[tex]c = -2-6[/tex]
[tex]c = -8[/tex]
[tex]f(x) = \frac{6}{x^2} + c[/tex] becomes
[tex]f(x) = \frac{6}{x^2} -8[/tex]
For: [tex]f'(x) = -\frac{12}{x^3}[/tex]
[tex]f(x) = \int {-\frac{12}{x^3} } \, dx[/tex]
Integrate:
[tex]f(x) = -\frac{12}{2x^2} + c[/tex]
[tex]f(x) = -\frac{6}{x^2} + c[/tex]
We understand that it passes through [tex](x,y) = (1,-2)[/tex].
So, we have:
[tex]-2 = -\frac{6}{1^2} + c[/tex]
[tex]-2 = -\frac{6}{1} + c[/tex]
[tex]-2 = -6 + c[/tex]
Make c the subject
[tex]c = -2+6[/tex]
[tex]c = 4[/tex]
[tex]f(x) = -\frac{6}{x^2} + c[/tex] becomes
[tex]f(x) = -\frac{6}{x^2} + 4[/tex]
You can use the formula for finding the arc length on specified interval on x axis.
The curves whose arc length on the given interval is described are
[tex]f(x) = 6x^{-2} -8[/tex]
and
[tex]f(x) = -6x^{-2} + 4[/tex]
If the function is differentiable in the given interval, then we have:
[tex]s = \int_a^b\sqrt{(1 + (f'(x))^2)}\:dx[/tex]
where s denotes the length of the arc of the given function from x = a to x = b
Using the above formula, as we're already given the arc length, thus,
[tex]\int_a^b\sqrt{(1 + (f'(x))^2)}\:dx = \int_2^6\sqrt{(1 +144x^{-6})}\:dx[/tex]
This gives us
[tex]f'(x) = \pm \sqrt{144x^{-6}} = \pm 12x^{-3}[/tex]
Integrating both sides with respect to x, we get:
[tex]f(x) = \pm \int 12x^{-3}\\\\f(x) = \pm 6x^{-2} + c[/tex]
where c is constant of integration.
Since the curve passes through (1,-2), thus, putting f(x) = -2, x = 1, we get:
[tex]f(x) = \pm 6x^{-2} + c\\-2 = \pm 6 + c\\c = -2 \mp 6 = -8, or ,4[/tex]
c = -8 when we have [tex]f(x) = 6x^{-2} + c[/tex]
c = + 4 when we have [tex]f(x) = -6x^{-2} + c[/tex]
The curves whose arc length on the given interval is described are
[tex]f(x) = 6x^{-2} -8[/tex]
and
[tex]f(x) = -6x^{-2} + 4[/tex]
Learn more about length of the arc of a curve here:
https://brainly.com/question/14319881
