1. With time, t, in minutes, the temperature, H, in degrees Celsius, of a bottle of water put in the refrigerator at time t=o is given by
H = 4 + 16e-.02t
How fast is the water cooling initially? After 10 minutes? Include units in your answer
To find how fast the water is cooling initially, we need to find the derivative of the temperature function with respect to time. The derivative will give us the rate at which the temperature is changing.
Given the temperature function:
H = 4 + 16e^(-0.02t)
To find the derivative, we can use the chain rule of derivatives. Let's start by differentiating the exponential term: e^(-0.02t).
d/dt(e^(-0.02t)) = -0.02 * e^(-0.02t)
Next, we differentiate the entire function H = 4 + 16e^(-0.02t) with respect to time:
dH/dt = 0 + 16 * (-0.02 * e^(-0.02t))
Simplifying further:
dH/dt = -0.32e^(-0.02t)
This expression represents the rate at which the water temperature is decreasing with respect to time. Since the derivative is with respect to minutes, the units for the rate of cooling are degrees Celsius per minute.
Now, to find how fast the water is cooling initially, we substitute t = 0 into the derivative expression:
dH/dt at t=0 = -0.32e^(-0.02 * 0)
= -0.32e^0
= -0.32
The initial rate of cooling is -0.32 degrees Celsius per minute.
To find the rate of cooling after 10 minutes, we substitute t = 10 into the derivative expression:
dH/dt at t=10 = -0.32e^(-0.02 * 10)
= -0.32e^(-0.2)
≈ -0.32 * 0.8187
≈ -0.2618
The rate of cooling after 10 minutes is approximately -0.2618 degrees Celsius per minute.
To find how fast the water is cooling initially, we need to find the rate of change of the temperature with respect to time at t = 0.
Differentiating the given equation with respect to t, we get:
dH/dt = d/dt(4 + 16e^(-0.02t))
Using the chain rule, we have:
dH/dt = 0 - 0.02 * 16 * e^(-0.02t)
At t = 0, the equation becomes:
dH/dt = -0.02 * 16 * e^(-0.02 * 0)
Since e^0 is equal to 1, the equation simplifies to:
dH/dt = -0.02 * 16 = -0.32
Therefore, initially, the water is cooling at a rate of -0.32 degrees Celsius per minute.
To find how fast the water is cooling after 10 minutes, we substitute t = 10 into the derivative equation:
dH/dt = -0.02 * 16 * e^(-0.02 * 10)
Using a calculator, we find:
dH/dt = -0.02 * 16 * e^(-0.2) ≈ -0.292
Therefore, after 10 minutes, the water is cooling at a rate of approximately -0.292 degrees Celsius per minute.