How To Find Arc In Math

How can one find the arc length of [math]y = \frac{x^2}{2} - \frac{\ln x}{4}[/math] from x = 2 to x = 4 without using a hyperbolic inverse sine for integration?

Therefore,the length equals 6.173286795...

How do you find the area of a cone? Gr 9 math I forgot lol?

The surface area of the cone is the area of the base added to the surface area of that top part. The base, of course, is simply [math]\pi r^2[/math], as the base is a circle with radius r. The area of the top part is trickier to find. If you cut a cone's pointy part so that you can lay that part flat, it will look like that little Pac-Man guy in this picture.(1)Now, that is simply a sector of a circle, and we know how to find the area of those! As the image shows, the arc length is [math]2 \pi r[/math], because that is just the circumference of the base of the cone (see the left part of the picture for verification). The radius of that sector is [math]\sqrt{r^2+h^2}[/math], where h is the height of the cone. The area of a sector of a circle is given by the formula[math]A=0.5 l r[/math], where l is the arc length and r is the radius of the sector. Plug the two values in, and we get [math]Surface Area= \pi r^2 + \pi r \sqrt{r^2+h^2}[/math](1) From onlinemathlearning.comI can do volume, too, if you want!

How do I find the arc length of [math]y = \sqrt{x-x^2} +\frac{\sqrt {x}}{\sin x}?[/math]

In general, the arc length for a continuous function [math]f(x)[/math] on [math]\text{[a,b]}[/math] is defined by [math]\int_a^b \sqrt{1+f'(x)^{2}} dx[/math].Your function simplifies to [math]y = \sqrt{x-x^{2}} + \sqrt{x}[/math]. The derivative [math]y'[/math] is needed in the arc length function. Using the power rule and chain rule (I’m too lazy to work everything out at 1 in the morning, so I let Wolfram Alpha do the thinking for me), we find that [math]y' = \frac{1 - 2 x}{2\sqrt{x - x^2}} - \frac{1}{2\sqrt{x}}[/math].Substituting this into the arc length formula gives [math]\int_a^b \sqrt{1+\left ( \frac{1 - 2 x}{2\sqrt{x - x^2}} - \frac{1}{2\sqrt{x}} \right )^{2}} dx[/math].Wolfram shows that there is no elementary way to integrate this function, so a numerical approximation sounds like it’s the best way to go. Now, the original function we want to find the arc length of is real-valued only on [math]\text[0,1][/math]. Therefore, the arc length of the original function over its entire real domain is[math]\int_0^1 \sqrt{1+\left ( \frac{1 - 2 x}{2\sqrt{x - x^2}} - \frac{1}{2\sqrt{x}} \right )^{2}} dx \approx[/math][math] 2.06076383150366[/math].Edit: Seeing that the original question was changed, I’ll solve it for the new equation.[math]y = \sqrt{x-x^{2}} + \frac{\sqrt{x}}{\sin{x}}[/math]In this case:[math]y' = \frac{1-2x}{2\sqrt{x-x^{2}}} + \frac{\csc{x}}{2\sqrt{x}} - \sqrt{x}\csc{x}\cot{x}[/math]The equation has an asymptote at [math]x=0[/math], so its arc length is actually infinite (undefined) over its entire real-valued domain. Regardless, you can still substitute [math]y'[/math] into the arc length formula and solve for it numerically, though you cannot finitely evaluate it with [math]a=0[/math].

How do you find the arc length of the parametric curve [math]x = \cos^3t, y = \sin^3t[/math] from [math]0[/math] to [math]2\pi[/math]? I get the correct answer if I quadruple the arc length from [math]0[/math] to [math]\frac{\pi}{2}[/math], but not if I try to evaluate the whole curve at once.

Once we make x and y easier to work with by noticing that:[math]y=sin^3t=\frac{1}{4}(3sin(t)-sin(3t))[/math][math]x=cos^3t=\frac{1}{4}(3cos(t)+cos(3t))[/math](using, for example sin(3x)=sin(2x+x) and then expanding)the derivatives with respect to t turn out to be:[math]\frac{\mathrm{d}y }{\mathrm{d} t}=\frac{1}{4}(3cos(t)-3cos(3t))[/math][math]\frac{\mathrm{d}x }{\mathrm{d} t}=-\frac{1}{4}(3sin(t)+3sin(3t))[/math]Then the arclength is:[math]S=\int_{0}^{2\pi}\sqrt{\frac{9}{16}(2+2(cos(t)cos(3t)+sin(t)sin(3t))}dt= \frac{3\sqrt2}{4}\int_{0}^{2\pi}\sqrt{1+cos(2t)}dt [/math](I did omit a lot of calculations, however I think that it’s unnecessary to show squaring the brackets and using the Pythagorean trigonometric identity, especially since OP has already solved the problem)While solving the problem, you have probably noticed that:[math]1+cos(2t)=2cos^2(t)[/math]Thus, the integral becomes:[math]\frac{3\sqrt2}{4}\int_{0}^{2\pi}\sqrt{2cos^2(t)}dt=\frac{3}{2}\int_{0}^{2\pi}|cos(t)|dt[/math]It is important to notice, that the sign of cos(t) changes , when t goes from [math]0\: [/math]to [math]2\pi\:[/math], and we wish to only add up the absolute value of the area under it, so the best way to solve the integral is to quadruple the arc length with the limits of [math]0\: [/math] to [math] \pi/2\:[/math], just as you did. So whatever different method you’re using to calculate the integral for the full range either forgets about the absolute value, or has a problem taking it into consideration, at least that’s my guess.And I think that[math]S=6[/math]

How can I find the arc distance from Havana to Madrid using math?

You need to a line segment between the two cities, Havana and Madrid. Assuming that the earth is round, you will need the length of the radius of the earth (distance from the surface to the center). From the center of the earth you will have a line segment going to Havana and another one going to Madrid. These 2 line segment form an angle. You need to know the angle (or an estimate) that these two line segments make. Once you have the angle and your radius use the following formula for your arc length, L.If θ is the degrees, then the length, L =( θ/360°) * 2 π r