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?
To find how fast the level of the milkshake is dropping, we need to find the rate at which the volume of the milkshake is changing with respect to time.
Let's start by finding the formula for the volume of the milkshake. Since the milkshake is in a cylindrical cup, the volume can be calculated using the formula for the volume of a cylinder:
V = πr^2h
Where:
V is the volume of the milkshake
π is a mathematical constant, approximately equal to 3.14159
r is the radius of the cup (5cm in this case)
h is the height (or level) of the milkshake
Since we are interested in finding how fast the level of the milkshake is dropping, we need to determine dh/dt (the rate of change of the height with respect to time).
Now, given that the student is drinking at a rate of 3.5cm^3 per second, we can assume that the rate at which the milkshake volume is changing is negative, indicating a decrease.
So, we can rewrite our formula for the volume of the milkshake as:
V = πr^2h(t)
Where h(t) represents the height of the milkshake as a function of time, and we want to find dh/dt when V = -3.5cm^3/s.
To find dh/dt, we can differentiate both sides of the formula with respect to time:
dV/dt = d(πr^2h(t))/dt
The left side of the equation represents the rate at which the volume is changing, which is -3.5cm^3/s. The right side of the equation represents the rate of change of the height, dh/dt, multiplied by the constant terms:
-3.5 = 2πr^2(dh/dt)
By rearranging the equation, we can solve for dh/dt:
dh/dt = (-3.5) / (2πr^2)
Plugging in the given radius (5cm) into the equation:
dh/dt = (-3.5) / (2π(5)^2)
Simplifying further:
dh/dt = (-3.5) / (2π(25))
Finally, we can calculate the rate at which the level of the milkshake is dropping:
dh/dt ≈ -0.0223 cm/s
Therefore, the level of the milkshake is dropping at a rate of approximately 0.0223 cm/s.