water is poured into a conical paper cup so that the height increases at the constant rate of 1 inch per second. if the cup is 6 inches tall and its top has a radius of 2 inches, how fast is the volume of water in the cup increasing when the height is 3 inches?
Make a diagram, and label the radius of the water level r, and the height of the water h
V = (1/3)πr^2h
but by ratios, r/h = 2/6 = 1/3
then r= h/3
V = (1/3)π(h^2/9)(h) = (π/27)h^3
dV/dt = (π/9)h^2 dh/dt
= (π/9)(3^2) (1) = π cubic inches/sec
check my arithmetic
To find how fast the volume of water in the cup is increasing, we need to use the formula for the volume of a cone:
V = (1/3)πr^2h,
where V is the volume, r is the radius of the top of the cone, and h is the height.
Given that the top radius (r) is 2 inches and the height (h) is increasing at a rate of 1 inch per second, we need to find the rate at which the volume (V) is changing when the height is 3 inches.
We can differentiate both sides of the volume formula with respect to time (t) to find the rate of change of the volume with respect to time:
dV/dt = (d(r^2)/dt * h + d(h)/dt * r^2).
Since the radius (r) is constant, its derivative with respect to time is zero (d(r^2)/dt = 0). And since the height (h) is increasing at a constant rate of 1 inch per second, its derivative d(h)/dt is also equal to 1.
Therefore, the equation simplifies to:
dV/dt = d(h)/dt * r^2,
dV/dt = 1 * (2^2).
dV/dt = 4 cubic inches per second.
So, when the height is 3 inches, the volume of water in the cup is increasing at a rate of 4 cubic inches per second.
To find how fast the volume of water in the cup is increasing, we need to use the rate at which the height is increasing and the given dimensions of the cup.
The volume of a cone is given by the formula V = (1/3) * π * r^2 * h, where V is the volume, π is a mathematical constant (approximately equal to 3.14159), r is the radius of the top of the cone, and h is the height of the cone.
Given:
- Rate of height increase = 1 inch per second
- Cup height (initial) = 6 inches
- Cup radius = 2 inches
- Height at which we want to find the rate of volume increase = 3 inches
First, let's find the radius of the cone at a height of 3 inches. We can use similar triangles to calculate this.
At a height of 6 inches, the radius is 2 inches.
So, the ratio of the radius to height is constant: r/h = 2/6 = 1/3.
Using this ratio, we can find the radius at a height of 3 inches: r = (1/3) * 3 = 1 inch.
Now that we have the radius and the height, we can calculate the volume of the cone at a height of 3 inches.
V = (1/3) * π * r^2 * h
= (1/3) * (3.14159) * (1^2) * 3
≈ 3.14159
So, the volume of water in the cone when the height is 3 inches is approximately 3.14159 cubic inches.
To find how fast the volume is increasing when the height is 3 inches, we can differentiate the equation for volume with respect to time (t).
dV/dt = (1/3) * π * (2 * r * dr/dt * h + r^2 * dh/dt)
Here, dr/dt represents the rate of change of the radius with respect to time, and dh/dt represents the rate of change of the height with respect to time. Since the height is increasing at a constant rate of 1 inch per second, dh/dt = 1.
We are interested in finding dV/dt when the height is 3 inches. At that moment, r = 1 inch, h = 3 inches, and dr/dt = 0 (since the radius is not changing over time).
dV/dt = (1/3) * π * (2 * 1 * 0 * 3 + 1^2 * 1)
= (1/3) * π * 1
≈ 1.0472
Therefore, the volume of water in the cup is increasing at a rate of approximately 1.0472 cubic inches per second when the height is 3 inches.