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Baking muffins: Chocolate muffins are baking in a 350 -degree oven. Let \(M=M(t)\) be the temperature of the muffins \(t\) minutes after they are placed in the oven, and let \(D=D(t)\) be the difference between the temperature of the oven and the temperature of the muffins. Then \(D\) satisfies Newton's law of cooling, and its equation of change is $$ \frac{d D}{d t}=-0.04 D $$ a. If the initial temperature of the muffins is room temperature, 73 degrees, find a formula for \(D\). b. Find a formula for \(M\). c. The muffins will be done when they reach a temperature of 225 degrees. When should we take the muffins out of the oven?

Short Answer

Expert verified
Bake muffins until approximately 25.1 minutes.

Step by step solution

01

Understand the Problem Terms

We have the following terms: - \(M(t)\): Temperature of the muffins at time \(t\).- \(D(t)\): Difference between the oven temperature and muffins' temperature.- Equation: \(\frac{d D}{d t}=-0.04 D\).- Initial muffins' temperature: 73 degrees (room temperature).- Oven temperature: Assumed constant at 350 degrees.
02

Solve for D(t) using Newton's Law

Newton's Law is given by \(\frac{d D}{d t}=-0.04 D\). This is a differential equation of the form \(\frac{d D}{D} = -0.04 \, dt\). Integrating both sides, we get: \[ \ln|D| = -0.04t + C \]Exponentiating both sides results in: \[ D(t) = D_0 e^{-0.04 t} \]where \(D_0\) is the initial difference when \(t = 0\). Given initial muffin temperature is 73 degrees: \[ D_0 = 350 - 73 = 277 \]Thus, \[ D(t) = 277 e^{-0.04 t} \].
03

Find M(t) in Terms of D(t)

Recall \(D(t) = 350 - M(t)\). We can express \(M(t)\) as: \[ M(t) = 350 - D(t) \]Substituting \(D(t)\) from Step 2, we have:\[ M(t) = 350 - 277 e^{-0.04 t} \].
04

Solve for Time When Muffins Reach 225 Degrees

We need to solve for \(t\) when \(M(t) = 225\). Substitute into \(M(t) = 350 - 277 e^{-0.04 t}\): \[ 225 = 350 - 277 e^{-0.04 t} \]Rearrange to form:\[ 277 e^{-0.04 t} = 125 \]Divide both sides by 277:\[ e^{-0.04 t} = \frac{125}{277} \]Taking natural logarithms:\[ -0.04 t = \ln \left(\frac{125}{277}\right) \]So, solve for \(t\):\[ t = -\frac{1}{0.04} \ln \left(\frac{125}{277}\right) \] Calculate:\[ t \approx 25.1 \text{ minutes} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
To understand how temperature changes over time, we use differential equations. These are mathematical equations that involve functions and their derivatives, showing how one factor changes in relation to another. In our muffin baking scenario, the differential equation represents how the difference in temperature (\(D(t)\)) between the muffins and the oven changes with time. The equation given, \( \frac{dD}{dt} = -0.04D \), is a simple type of differential equation known as a "separable" equation. Here, it implies that the rate of change of the temperature difference is proportional to the current difference in temperature. This is typical in processes involving heating or cooling under controlled temperature conditions. Such equations are crucial in accurately predicting how quickly or slowly temperature changes, which is essential for understanding the physics of cooling or heating objects. By integrating both sides of this equation, we can solve for the function \(D(t)\), which gives us the temperature difference as a function of time.
Exponential Functions
Exponential functions are a key component when dealing with differential equations like the one in our muffin scenario. Once you integrate a differential equation that has a form similar to \( \frac{dD}{dt} = -kD \), where \(k\) is a constant, you land on an exponential function. In this case, after solving the differential equation, we obtain \( D(t) = D_0 e^{-0.04t} \). Here, \(e\) is the base of the natural logarithm, an important constant in mathematics, while \(D_0\) is the initial temperature difference of 277 degrees. The exponential nature of this function tells us that the temperature difference decreases exponentially over time, meaning it decreases rapidly initially, and then more slowly as it approaches zero. Using exponential functions simplifies the prediction of when the muffins will reach a certain temperature, as these functions align well with natural decay or growth processes, making them invaluable for scenarios involving cooling or heating.
Modeling Real-life Scenarios
Modeling real-life scenarios mathematically helps us apply theoretical concepts to practical situations, like baking muffins. Newton's Law of Cooling is a prime example. By using a differential equation, we can simulate how muffins heat up to reach a specific temperature, which is critical for perfectly baked pastries. Our equation, along with the initial conditions provided, allows us to develop a mathematical model for the muffins' temperature change over time. This model can predict exactly when the muffins have reached a temperature of 225 degrees, using the function \( M(t) = 350 - 277 e^{-0.04t} \). With this method, not only do we learn about the theoretical framework—using calculus and exponential functions—but we also gain the capability to apply these concepts to everyday activities. This cross-over of theory into practical applications exemplifies the power of mathematical modeling in understanding and predicting real-world phenomena such as cooking or climate control.

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Most popular questions from this chapter

Radioactive decay: The amount remaining \(A\), in grams, of a radioactive substance is a function of time \(t\), measured in days since the experiment began. The equation of change for \(A\) is $$ \frac{d A}{d t}=-0.05 A $$ a. What is the exponential growth rate for \(A\) ? b. If initially there are 3 grams of the substance, find a formula for \(A\). c. What is the half-life of this radioactive substance?

Falling with a parachute: When an average-size man with a parachute jumps from an airplane, he will fall \(S=12.5\left(0.2^{t}-1\right)+20 t\) feet in \(t \mathrm{sec}-\) onds. a. Plot the graph of \(S\) versus \(t\) over at least the first 10 seconds of the fall. b. How far does the parachutist fall in 2 seconds? c. Calculate \(\frac{d S}{d t}\) at 2 seconds into the fall and explain what the number you calculated means in practical terms.

Looking up: The constant \(g=32\) feet per second per second is the downward acceleration due to gravity near the surface of the Earth. If we stand on the surface of the Earth and locate objects using their distance up from the ground, then the positive direction is up, so down is the negative direction. With this perspective, the equation of change in velocity for a freely falling object would be expressed as $$ \frac{d V}{d t}=-g $$ (We measure upward velocity \(V\) in feet per second and time \(t\) in seconds.) Consider a rock tossed up- ward from the surface of the Earth with an initial velocity of 40 feet per second upward. a. Use a formula to express the velocity \(V=V(t)\) as a linear function. (Hint: You get the slope of \(V\) from the equation of change. The vertical intercept is the initial value.) b. How many seconds after the toss does the rock reach the peak of its flight? (Hint: What is the velocity of the rock when it reaches its peak?) c. How many seconds after the toss does the rock strike the ground? (Hint: How does the time it takes for the rock to rise to its peak compare with the time it takes for it to fall back to the ground?)

Mileage for an old car: The gas mileage \(M\) that you get on your car depends on its age \(t\) in years. a. Explain the meaning of \(\frac{d M}{d t}\) in practical terms. b. As your car ages and performance degrades, do you expect \(\frac{d M}{d t}\) to be positive or negative?

Experimental determination of the drag coefficient: When retardation due to air resistance is proportional to downward velocity \(V\), in feet per second, falling objects obey the equation of change $$ \frac{d V}{d t}=32-r V $$ where \(r\) is known as the drag coefficient. One way to measure the drag coefficient is to measure and record terminal velocity. a. We know that an average-size man has a terminal velocity of 176 feet per second. Use this to show that the value of the drag coefficient is \(r=0.1818\) per second. (Hint: To say that the terminal velocity is 176 feet per second means that when the velocity \(V\) is 176 , velocity will not change. That is, \(\frac{d V}{d t}=0\). Put these bits of information into the equation of change and solve for \(r\).) b. An ordinary coffee filter has a terminal velocity of about 4 feet per second. What is the drag coefficient for a coffee filter?

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