sand is being dropped onto a conical pile in such a way that the height of the pile is always twice the base radius.what is the rate of change of the volume of the pile with respect to the radius when the radius of the pile is 12 inches?
you need dV/dr
V = (1/3) pi r^2 (2r) = (2/3) pi r^3
dV/dr = (2/3)pi * 3 r^2
= 2 pi r^2
when r = 12
dV/dr = 288 pi
Ah, we're talking about sand and cones today, huh? Well, let's dive right in, but don't worry, I promise it won't be as messy as a clown juggling sand.
So, we have a conical pile where the height is always twice the base radius. That means if the radius (let's call it r) of the pile is 12 inches, then the height (let's call it h) would be 2 times that, or 24 inches.
The volume (V) of a cone is given by the formula V = (1/3)πr²h, where π is a mathematical constant.
Now, to find the rate of change of the volume with respect to the radius (dr/dV), we take the derivative of the volume equation with respect to r.
dV/dr = (1/3)π * (2rh)
Substituting r = 12 and h = 24, we get:
dV/dr = (1/3)π * (2 * 12 * 24)
And after some quick calculations, we find that the rate of change of the volume with respect to the radius when the radius is 12 inches is:
dV/dr ≈ 301.59 cubic inches per inch
So, the volume of the pile would change by approximately 301.59 cubic inches for every inch increase in radius.
Hope that helps! If not, don't worry, I still have plenty of clown tricks up my sleeve.
To find the rate of change of the volume of the pile with respect to the radius, we need to differentiate the volume with respect to the radius and then evaluate it at a given radius.
First, let's determine the formula for the volume of a cone. The volume of a cone (V) is given by the formula:
V = (1/3) * π * r^2 * h,
where r is the radius and h is the height of the cone.
Given that the height of the pile is always twice the base radius, we can express the height as h = 2r.
Now, substitute h = 2r into the volume formula and simplify:
V = (1/3) * π * r^2 * (2r)
= (2/3) * π * r^3.
Taking the derivative of the volume with respect to the radius (r), we get:
dV/dr = d/dt [(2/3) * π * r^3]
= (2/3) * π * 3r^2
= 2πr^2.
To find the rate of change of the volume with respect to the radius when the radius of the pile is 12 inches (r = 12), substitute r = 12 into the derivative equation:
dV/dr = 2π(12)^2
= 2π(144)
= 288π in^2.
Therefore, the rate of change of the volume of the pile with respect to the radius when the radius is 12 inches is 288π cubic inches per inch.
To find the rate of change of the volume of the pile with respect to the radius, we need to differentiate the volume formula of the conical pile with respect to the radius and then substitute the given value for the radius.
Let's go step by step:
1. The volume formula for a cone is given by V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
2. In this case, the height of the cone is always twice the base radius, so we can write h = 2r.
3. Substituting the value of h in the volume formula, we get V = (1/3)πr^2(2r) = (2/3)πr^3.
4. Now, we need to differentiate the volume formula with respect to the radius (r). To do this, we can apply the power rule of differentiation, which states that d/dx(x^n) = nx^(n-1). In our case, n = 3, so we differentiate (2/3)πr^3 as follows:
dV/dr = d/dx((2/3)πx^3) = (2/3)π * 3r^(3-1) = (2/3)π * 3r^2 = 2πr^2.
5. Now, we have the derivative of the volume with respect to the radius: dV/dr = 2πr^2.
6. To find the rate of change of the volume with respect to the radius when the radius is 12 inches, we substitute the given value into the derivative: dV/dr = 2π(12^2) = 2π(144) = 288π.
Therefore, the rate of change of the volume of the pile with respect to the radius when the radius is 12 inches is 288π cubic inches per inch.