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[https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_2#Solution_5 2024 AIME I Problem 2] | [https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_2#Solution_5 2024 AIME I Problem 2] | ||
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+ | [https://artofproblemsolving.com/wiki/index.php/2024_AMC_10B_Problems/Problem_10#Solution_4 2024 AMC 10B Problem 10] | ||
==Some Proofs I wrote== | ==Some Proofs I wrote== | ||
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===Volume of Cylinder, Cone, and Sphere=== | ===Volume of Cylinder, Cone, and Sphere=== | ||
− | If we have a function <math>f(x)</math>, that can be rotated to make a shape, the area underneath it will turn into the volume. However, since we are revolving it in a circular motion, the area will actually become the radius. Another way of seeing this is splitting it into infinite circles and adding up all of them. Therefore, for a function <math>f(x)</math>, we have the volume of the solid of revolution to be < | + | If we have a function <math>f(x)</math>, that can be rotated to make a shape, the area underneath it will turn into the volume. However, since we are revolving it in a circular motion, the area will actually become the radius. Another way of seeing this is splitting it into infinite circles and adding up all of them. Therefore, for a function <math>f(x)</math>, we have the volume of the solid of revolution to be <cmath>\pi \int_{a}^{b} (f(x))^2 \,dx </cmath>. |
− | Cylinder: A cylinder can be expressed a solid of revolution by revolving the line < | + | Cylinder: A cylinder can be expressed a solid of revolution by revolving the line <math>y = r</math> around the <math>x</math>-axis. To find the volume, we can find the area under the curve, and then when we revolve it, it becomes the volume. The radius is <math>r</math> and the height, <math>h</math>, is the upper bound of integration. We have <cmath>\pi \int_{0}^{h} r^2 \,dx </cmath>. Integrating, we get <cmath>\pi (r^2h - r^2(0)) = \pi r^2h</cmath>. This is the formula of a cylinder. |
− | Cone: If you are given the height and radius of the cone, and you have the point (0,0) on your line(since the vertex is 0), then < | + | Cone: If you are given the height and radius of the cone, and you have the point <math>(0,0)</math> on your line(since the vertex is 0), then <math>f(h) = r</math>, because the height is the x-coordinate and the radius is the y(for the same reason seen above in the cylinder). Now, since we have <math>(0,0)</math>, we know the y-intercept, and we can only have one slope. If <math>h=x</math>, and <math>m</math> is the slope, then we have <math>r = mh</math>, and therefore <math>m = \frac{r}{h}</math>, so the equation is <math>f(x) = \frac{rx}{h}</math>. For the integral, we get <cmath>\pi \int_{0}^{h} \frac{r^2x^2}{h^2} \,dx = \pi \frac{r^2h}{3}</cmath>. |
− | Sphere: The equation of a sphere should be a circle, but that is a relation and not a function. Therefore, we can use the top half of a circle, and the bottom half will get filled in when it rotates. Therefore, we get < | + | Sphere: The equation of a sphere should be a circle, but that is a relation and not a function. Therefore, we can use the top half of a circle, and the bottom half will get filled in when it rotates. Therefore, we get <math>f(x) = \sqrt{r^2 - x^2}</math>. The diameter is <math>-r</math> to <math>r</math>, so that is where we integrate. <cmath>\pi \int_{-r}^{r} r^2 - x^2 \,dx = \frac{4\pi r^3}{3}</cmath>. |
Latest revision as of 12:38, 15 November 2024
Contents
[hide]My Solutions
Some Proofs I wrote
if is prime.
Proof: Expanding out, all the coefficients are of the form by the binomial theorem. To prove the original result we must show that if and , then . Because , , which is divisible by , so the original expression must be divisible by . However if is prime, , since does not contain (because ). Therefore, in order for to be divisible by , is divisible by . All the coefficients of the expansion(besides the coefficients of and ) are of the form , and , so they cancel out and if is prime.
Volume of Cylinder, Cone, and Sphere
If we have a function , that can be rotated to make a shape, the area underneath it will turn into the volume. However, since we are revolving it in a circular motion, the area will actually become the radius. Another way of seeing this is splitting it into infinite circles and adding up all of them. Therefore, for a function , we have the volume of the solid of revolution to be .
Cylinder: A cylinder can be expressed a solid of revolution by revolving the line around the -axis. To find the volume, we can find the area under the curve, and then when we revolve it, it becomes the volume. The radius is and the height, , is the upper bound of integration. We have . Integrating, we get . This is the formula of a cylinder.
Cone: If you are given the height and radius of the cone, and you have the point on your line(since the vertex is 0), then , because the height is the x-coordinate and the radius is the y(for the same reason seen above in the cylinder). Now, since we have , we know the y-intercept, and we can only have one slope. If , and is the slope, then we have , and therefore , so the equation is . For the integral, we get .
Sphere: The equation of a sphere should be a circle, but that is a relation and not a function. Therefore, we can use the top half of a circle, and the bottom half will get filled in when it rotates. Therefore, we get . The diameter is to , so that is where we integrate. .