Math / Calculus

QuestionIf a cup of coffee has temperature $91^{\circ} \mathrm{C}$ in a room where the ambient air temperature is $19^{\circ} \mathrm{C}$ , then, according to Newton's Law of Cooling, the temperature of the coffee after $t$ minutes is $T(t)=19+72 e^{-t / 48}$ . What is the average temperature of the coffee during the first 16 minutes? average temp = $\square$ 91 ${ }^{\circ} \mathrm{C}$

Studdy Solution

STEP 1

What is this asking? Find the average temperature of a cup of coffee over the first 16 minutes, given a formula that tells us the temperature at any given minute. Watch out! Don't forget that we're looking for the average temperature, not just the temperature at t = 16 minutes!

STEP 2

1. Set up the average temperature integral.
2. Evaluate the integral.
3. Calculate the average temperature.

STEP 3

Alright, so we want the average temperature over the first 16 minutes.
Remember, the average value of a function $f(x)$ over an interval $[a, b]$ is given by $\frac{1}{b-a} \int_{a}^{b} f(x) \, dx$ .
This formula is like magic!
It spreads out all the temperatures over those 16 minutes and gives us the perfect average.

STEP 4

In our case, $f(x)$ is actually $T(t)$ , our temperature function $T(t) = 19 + 72e^{-t/48}$ .
Our interval starts at $t = 0$ minutes and ends at $t = 16$ minutes, so $a = 0$ and $b = 16$ .

STEP 5

Let's plug everything into our magical average value formula: $\frac{1}{16 - 0} \int_{0}^{16} (19 + 72e^{-t/48}) \, dt$ This integral represents the area under the temperature curve divided by the time interval, which gives us the average temperature.

STEP 6

Let's break down this integral into smaller, bite-sized pieces!
We can split it into two separate integrals: $\frac{1}{16} \left[ \int_{0}^{16} 19 \, dt + \int_{0}^{16} 72e^{-t/48} \, dt \right]$

STEP 7

The first integral is easy peasy: $\int_{0}^{16} 19 \, dt = 19t \Big|_0^{16} = 19 \cdot 16 - 19 \cdot 0 = \textbf{304}$

STEP 8

Now, for the second integral, we can use a u-substitution.
Let $u = -t/48$ , so $du = -dt/48$ or $dt = -48 \, du$ .
When $t = 0$ , $u = 0$ , and when $t = 16$ , $u = -16/48 = \textbf{-1/3}$ . $\int_{0}^{16} 72e^{-t/48} \, dt = 72 \int_{0}^{-1/3} e^u (-48) \, du = -3456 \int_{0}^{-1/3} e^u \, du$

STEP 9

Evaluating the integral, we get: $-3456 \left[ e^u \Big|_0^{-1/3} \right] = -3456 (e^{-1/3} - e^0) = -3456(e^{-1/3} - 1) = 3456(1 - e^{-1/3})$ This simplifies to approximately 989.64.

STEP 10

Putting it all together: $\frac{1}{16} [304 + 989.64] = \frac{1}{16} [1293.64]$

STEP 11

Finally, let's calculate that average temperature: $\frac{1293.64}{16} \approx \textbf{80.85}$

STEP 12

The average temperature of the coffee during the first 16 minutes is approximately $80.85^{\circ} \mathrm{C}$ .

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Studdy Solution

STEP 1

What is this asking? Find the average temperature of a cup of coffee over the first 16 minutes, given a formula that tells us the temperature at any given minute. Watch out! Don't forget that we're looking for the average temperature, not just the temperature at t = 16 minutes!

STEP 2

1. Set up the average temperature integral.
2. Evaluate the integral.
3. Calculate the average temperature.

STEP 3

STEP 4

In our case, $f(x)$ is actually $T(t)$ , our temperature function $T(t) = 19 + 72e^{-t/48}$ .
Our interval starts at $t = 0$ minutes and ends at $t = 16$ minutes, so $a = 0$ and $b = 16$ .

STEP 5

Let's plug everything into our magical average value formula: $\frac{1}{16 - 0} \int_{0}^{16} (19 + 72e^{-t/48}) \, dt$ This integral represents the area under the temperature curve divided by the time interval, which gives us the average temperature.

STEP 6

Let's break down this integral into smaller, bite-sized pieces!
We can split it into two separate integrals: $\frac{1}{16} \left[ \int_{0}^{16} 19 \, dt + \int_{0}^{16} 72e^{-t/48} \, dt \right]$

STEP 7

The first integral is easy peasy: $\int_{0}^{16} 19 \, dt = 19t \Big|_0^{16} = 19 \cdot 16 - 19 \cdot 0 = \textbf{304}$

STEP 8

Now, for the second integral, we can use a u-substitution.
Let $u = -t/48$ , so $du = -dt/48$ or $dt = -48 \, du$ .
When $t = 0$ , $u = 0$ , and when $t = 16$ , $u = -16/48 = \textbf{-1/3}$ . $\int_{0}^{16} 72e^{-t/48} \, dt = 72 \int_{0}^{-1/3} e^u (-48) \, du = -3456 \int_{0}^{-1/3} e^u \, du$

STEP 9

Evaluating the integral, we get: $-3456 \left[ e^u \Big|_0^{-1/3} \right] = -3456 (e^{-1/3} - e^0) = -3456(e^{-1/3} - 1) = 3456(1 - e^{-1/3})$ This simplifies to approximately 989.64.

STEP 10

Putting it all together: $\frac{1}{16} [304 + 989.64] = \frac{1}{16} [1293.64]$

STEP 11

Finally, let's calculate that average temperature: $\frac{1293.64}{16} \approx \textbf{80.85}$

STEP 12

The average temperature of the coffee during the first 16 minutes is approximately $80.85^{\circ} \mathrm{C}$ .

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Studdy Solution

STEP 1

What is this asking? Find the average temperature of a cup of coffee over the first 16 minutes, given a formula that tells us the temperature at any given minute. Watch out! Don't forget that we're looking for the *average* temperature, not just the temperature at *t* = 16 minutes!

STEP 2

1. Set up the average temperature integral.2. Evaluate the integral.3. Calculate the average temperature.

STEP 3

STEP 4

In our case, f(x)f(x)f(x) is actually T(t)T(t)T(t), our temperature function T(t)=19+72e−t/48T(t) = 19 + 72e^{-t/48}T(t)=19+72e−t/48.Our interval starts at t=0t = 0t=0 minutes and ends at t=16t = 16t=16 minutes, so a=0a = 0a=0 and b=16b = 16b=16.

STEP 5

STEP 6

Let's break down this integral into smaller, bite-sized pieces!We can split it into two separate integrals: 116[∫01619 dt+∫01672e−t/48 dt] \frac{1}{16} \left[ \int_{0}^{16} 19 \, dt + \int_{0}^{16} 72e^{-t/48} \, dt \right] 161​[∫016​19dt+∫016​72e−t/48dt]

STEP 7

The first integral is easy peasy: ∫01619 dt=19t∣016=19⋅16−19⋅0=304 \int_{0}^{16} 19 \, dt = 19t \Big|_0^{16} = 19 \cdot 16 - 19 \cdot 0 = \textbf{304} ∫016​19dt=19t∣∣​016​=19⋅16−19⋅0=304

STEP 8

STEP 9

STEP 10

Putting it all together: 116[304+989.64]=116[1293.64] \frac{1}{16} [304 + 989.64] = \frac{1}{16} [1293.64] 161​[304+989.64]=161​[1293.64]

STEP 11

Finally, let's calculate that average temperature: 1293.6416≈80.85 \frac{1293.64}{16} \approx \textbf{80.85} 161293.64​≈80.85

STEP 12

The average temperature of the coffee during the first 16 minutes is approximately 80.85∘C80.85^{\circ} \mathrm{C}80.85∘C.

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Start learning now

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What is this asking? Find the average temperature of a cup of coffee over the first 16 minutes, given a formula that tells us the temperature at any given minute. Watch out! Don't forget that we're looking for the average temperature, not just the temperature at t = 16 minutes!

1. Set up the average temperature integral.
2. Evaluate the integral.
3. Calculate the average temperature.

In our case, $f(x)$ is actually $T(t)$ , our temperature function $T(t) = 19 + 72e^{-t/48}$ .
Our interval starts at $t = 0$ minutes and ends at $t = 16$ minutes, so $a = 0$ and $b = 16$ .

Let's break down this integral into smaller, bite-sized pieces!
We can split it into two separate integrals: $\frac{1}{16} \left[ \int_{0}^{16} 19 \, dt + \int_{0}^{16} 72e^{-t/48} \, dt \right]$

The first integral is easy peasy: $\int_{0}^{16} 19 \, dt = 19t \Big|_0^{16} = 19 \cdot 16 - 19 \cdot 0 = \textbf{304}$

Putting it all together: $\frac{1}{16} [304 + 989.64] = \frac{1}{16} [1293.64]$

Finally, let's calculate that average temperature: $\frac{1293.64}{16} \approx \textbf{80.85}$

The average temperature of the coffee during the first 16 minutes is approximately $80.85^{\circ} \mathrm{C}$ .