A student is drinking a milkshake with a straw from a cylindrical cup with a radius of 5cm. if the student is drinking at a rate of 3.5cm^3 per second, how fast is the level of a milkshake dropping?
dv/dt = pi r^2 dh/dt
just plug in your numbers to find dh/dt
let the height of the shake level be h cm
V= πr^2 h
= 25πh
dV/dt = 25π dh/dt
-3.5 = 25π dh/dt
dh/dt = -3.5/25π = appr -.04456.. cm^3/s
To find the rate at which the level of the milkshake is dropping, we need to use the formula for the volume of a cylinder.
The formula for the volume of a cylinder is V = πr^2h, where V is the volume, r is the radius, and h is the height (or level) of the cylinder.
In this case, the radius (r) is given as 5 cm, and we are given the rate at which the student is drinking, which is 3.5 cm^3/s. We want to find how fast the level (h) is dropping.
To find the rate at which the level is dropping, we need to differentiate the volume formula with respect to time (t):
dV/dt = d/dt (πr^2h)
Now, let's differentiate both sides of the equation:
dV/dt = 2πrh * dh/dt
We can rearrange the equation to solve for dh/dt:
dh/dt = (dV/dt) / (2πrh)
Now we can substitute the given values into the equation:
dh/dt = (3.5 cm^3/s) / (2π(5 cm)(5 cm))
Calculating this, we get:
dh/dt ≈ 0.0445 cm/s
Therefore, the level of the milkshake is dropping at a rate of approximately 0.0445 cm/s.
To determine how fast the level of the milkshake is dropping, we need to calculate the rate at which the volume of the milkshake is decreasing.
The volume of a cylinder can be calculated using the formula V = πr^2h, where V is the volume, r is the radius, and h is the height.
In this case, we are interested in finding the rate at which the height (level) of the milkshake is changing over time. So, we need to differentiate the volume formula with respect to time t:
dV/dt = d(πr^2h)/dt
To find how fast the level of the milkshake is dropping (dh/dt), we need to isolate this term in the above equation.
First, let's differentiate the volume formula:
dV/dt = 2πrh(dr/dt) + πr^2(dh/dt)
Now, let's manipulate the equation to solve for dh/dt:
dV/dt = 2πrh(dr/dt) + πr^2(dh/dt)
dh/dt (πr^2) = dV/dt - 2πrh(dr/dt)
dh/dt = (dV/dt - 2πrh(dr/dt))/(πr^2)
Given that the rate of drinking is 3.5 cm^3/s, we can substitute dV/dt with -3.5 cm^3/s (negative sign indicates volume is decreasing). Also, since the radius r is 5cm, we have:
dh/dt = (-3.5 - 2π(5)(3.5))/(π(5)^2)
Simplifying the equation:
dh/dt = (-3.5 - 2π(5)(3.5))/(25π)
dh/dt = (-3.5 - 35π)/(25π)
Finally, calculating the approximate value:
dh/dt ≈ (-3.5 - 35×3.14)/(25×3.14)
dh/dt ≈ (-3.5 - 109.9)/(78.5)
dh/dt ≈ -113.4/78.5
dh/dt ≈ -1.44 cm/s
Therefore, the level of the milkshake is dropping at a rate of approximately -1.44 cm/s. The negative sign indicates that the level is decreasing.