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
Do you have a typo?
Should it not be H = 4 + 16e^(-.02t) ?
if so, then ....
dH/dt = - .32e^(-.02t)
Initially means t = 0
dH/dt = -.32e^0 = - .32° / minute
when t=10
dH/dt = -.32e^(-.2) = -.26° / min
check my arithmetic
Well, when it comes to cooling water, things can get pretty cool! Let's find out how fast the water is cooling initially and after 10 minutes using some mathematical magic!
To determine the initial cooling rate, we need to calculate the derivative of the temperature equation with respect to time (t). So, let's put on our mathematical magician hats and get started!
The function we have is H = 4 + 16e^(-0.02t). To find the derivative, we can use the chain rule. So, let's hold our breath and do some calculations!
The derivative of H with respect to t is given by:
dH/dt = -0.02(16e^(-0.02t))
Now, let's find out how fast the water is cooling initially when t = 0. Plugging in 0 for t:
dH/dt = -0.02(16e^(-0.02*0)) = -0.02(16e^0) = -0.02(16) = -0.32
Voila! The water is cooling at a rate of -0.32 degrees Celsius per minute initially.
What about after 10 minutes, you ask? Let's use our trusty formula to find that out!
dH/dt = -0.02(16e^(-0.02t))
Plugging in 10 for t:
dH/dt = -0.02(16e^(-0.02*10)) = -0.02(16e^(-0.2)) ≈ -0.02(16 * 0.8187) ≈ -0.524 degrees Celsius per minute
So, after 10 minutes, the water is cooling at a rate of approximately -0.524 degrees Celsius per minute.
Stay cool, my friend, and keep those math questions coming!
To find how fast the water is cooling initially, we need to find the derivative of the temperature formula with respect to time, t, and evaluate it at time t=0.
The derivative of H with respect to t can be found using the chain rule:
dH/dt = (-0.02)(16)e^(-0.02t)
Plugging in t=0, we get:
dH/dt = (-0.02)(16)e^(-0.02(0))
= (-0.02)(16)e^0
= -0.02(16)(1)
= -0.32
Therefore, the water is cooling initially at a rate of -0.32 degrees Celsius per minute.
To find how fast the water is cooling after 10 minutes, we again find the derivative and evaluate it at t=10.
dH/dt = (-0.02)(16)e^(-0.02t)
Plugging in t=10, we get:
dH/dt = (-0.02)(16)e^(-0.02(10))
= (-0.02)(16)e^(-0.2)
≈ -0.0646
Therefore, the water is cooling at a rate of approximately -0.0646 degrees Celsius per minute after 10 minutes.
Note: The negative sign indicates that the temperature is decreasing.
To find how fast the water is cooling, we need to calculate its rate of change with respect to time, also known as the derivative. In this case, we need to find the derivative of the temperature function H with respect to time (t).
The given temperature function is:
H = 4 + 16e^(-0.02t)
To find the derivative, we can use the chain rule of differentiation:
Step 1: Differentiate the constant term:
The derivative of 4 with respect to t is 0 since it is a constant.
Step 2: Differentiate the exponential term:
The derivative of e^(-0.02t) with respect to t is (-0.02)e^(-0.02t), using the chain rule.
Step 3: Multiply the derivative of the exponential term by the constant term:
The derivative of H with respect to t is given by:
dH/dt = 0 + (16)(-0.02)e^(-0.02t)
Simplifying further:
dH/dt = -0.32e^(-0.02t)
Now we have the derivative of H with respect to t.
To find the initial cooling rate, we substitute t = 0 into the derivative equation:
dH/dt = -0.32e^(-0.02 * 0)
dH/dt = -0.32e^(0)
dH/dt = -0.32(1)
dH/dt = -0.32
Therefore, the water is cooling initially at a rate of -0.32 degrees Celsius per minute.
To find the cooling rate after 10 minutes, we substitute t = 10 into the derivative equation:
dH/dt = -0.32e^(-0.02 * 10)
dH/dt = -0.32e^(-0.2)
dH/dt ≈ -0.32(0.8187)
dH/dt ≈ -0.2622
Therefore, after 10 minutes, the water is cooling at a rate of approximately -0.2622 degrees Celsius per minute.
Note: The units for the cooling rate are degrees Celsius per minute.