User:Icepinner/LATEX practice

${\textstyle {\frac {2}{3}}}$ $\sin {\theta }$

${\textstyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

The formula for the volume of a sphere is ${\textstyle V={\frac {4}{3}}\pi r^{3}}$, where ${\textstyle r}$ is the radius of the sphere

The form for a spiral may be expressed as ${\textstyle r=a+b\theta ^{\frac {1}{c}}(c\neq 0}$), though an Archimedean spiral only occurs when ${\textstyle c=1}$

${\textstyle h'(x)=f(x)g'(x)+f'(x)g(x)}$

${\textstyle F=ma}$

${\textstyle E_{p}={\frac {GMm}{r}}}$

For distance of closest approach: ${\textstyle E_{k}i+E_{p}i=E_{k}f+E_{p}f}$

As the electric potential energy between an alpha particle and the nucleus of a gold atom is 0 when the alpha particle is at its starting position, ${\textstyle E_{k}i}$ (${\textstyle E_{k}i={\frac {1}{2}}mv^{2}}$) will be at its maximum. However, as the alpha particle approaches the nucleus, it will be deflected, indicating ${\textstyle E_{p}f}$ (${\textstyle E_{p}f={\frac {kQq}{r^{2}}}}$) will be at its maximum. Assuming energy is not lost, we can use the conservation of energy to establish that ${\textstyle E_{k}i=E_{p}f}$, thereby implying that ${\textstyle {\frac {1}{2}}mv^{2}={\frac {kQq}{r^{2}}}}$. Rearranging this relationship for ${\textstyle r}$ will yield ${\textstyle r={\sqrt {\frac {2KQq}{mv^{2}}}}}$, the distance of closest approach between the alpha particle and nucleus

Derivation for the volume of a cone

We have established that the radius of the cone, ${\textstyle r}$, and the height, ${\textstyle h}$, form a right angle triangle with the cone's slant height. We can represent this in the form of a linear function (or straight line) ${\textstyle f(x)=mx+b}$. If one were to treat the x-axis as the height and the y-axis as the radius, then they can represent said function as ${\textstyle f(x)={\frac {r}{h}}x+r}$. To obtain the volume of the cone's formula, we can rotate the function – from the function's x-intercept, ${\textstyle (-h,0)}$, to a value of ${\textstyle h}$ dictated by ${\textstyle r}$ – along the x-axis by ${\textstyle 2\pi }$ radians through solid of revolution and apply the volume of revolution formula ${\textstyle V=\pi \int _{a}^{b}f(x)^{2}dx}$. This yields the following:

${\textstyle V=\pi \int _{-h}^{h}({\frac {r}{h}}x+r)^{2}dx}$
${\textstyle V=\pi \int _{-h}^{h}{\frac {r^{2}}{h^{2}}}x^{2}+2{\frac {r^{2}}{h}}x+r^{2}dx}$
${\textstyle V=\pi \left[{\frac {1}{3}}{\frac {r^{2}}{h^{2}}}x^{3}+{\frac {r^{2}}{h}}x^{2}+r^{2}x\right]_{-h}^{h}}$
${\textstyle V=\pi [({\frac {1}{3}}{\frac {r^{2}}{h^{2}}}h^{3}+{\frac {r^{2}}{h}}h^{2}+r^{2}h)-(-{\frac {1}{3}}{\frac {r^{2}}{h^{2}}}h^{3}+{\frac {r^{2}}{h}}h^{2}-r^{2}h)]}$
${\textstyle V=\pi ({\frac {2}{3}}r^{2}h+2r^{2}h)}$