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The concentration \(C,\) in parts per million, of a medication in the body \(t\) hours after ingestion is given by the function \(C(t)=10 t^{2} e^{-t}\) a) Find the concentration after \(0 \mathrm{hr}, 1 \mathrm{hr}, 2 \mathrm{hr}, 3 \mathrm{hr},\) and \(10 \mathrm{hr}\). b) Sketch a graph of the function for \(0 \leq t \leq 10\). c) Find the rate of change of the concentration, \(C^{\prime}(t)\). d) Find the maximum value of the concentration and the time at which it occurs. e) Interpret the meaning of the derivative.

Short Answer

Expert verified
a) Concentrations are 0, 3.68, 5.41, 4.48, 0.045. b) Sketch shows rise and fall of concentration. c) Derivative is \(C'(t) = 10e^{-t}(2t - t^2)\). d) Maximum is 5.41 at 2 hr. e) Derivative shows rate and direction of concentration change.

Step by step solution

01

Calculate Concentration at Specific Times

Use the function \(C(t) = 10t^2 e^{-t}\) to find the concentration at \(t = 0, 1, 2, 3,\) and \(10\) hours by plugging each value into the function.- For \(t = 0\), \(C(0) = 10(0)^2 e^{0} = 0\).- For \(t = 1\), \(C(1) = 10(1)^2 e^{-1} \approx 3.68\).- For \(t = 2\), \(C(2) = 10(2)^2 e^{-2} \approx 5.41\).- For \(t = 3\), \(C(3) = 10(3)^2 e^{-3} \approx 4.48\).- For \(t = 10\), \(C(10) = 10(10)^2 e^{-10} \approx 0.045\).
02

Sketch the Graph of the Function

Plot the points calculated in Step 1: \((0,0), (1,3.68), (2,5.41), (3,4.48)\), and \((10,0.045)\). Draw a smooth curve through the points to represent the function over \(0 \leq t \leq 10\), showing an initial rise in concentration followed by a decrease.
03

Find the Derivative of the Function

To find the rate of change of the concentration, compute the derivative \(C'(t)\):\[ C'(t) = \frac{d}{dt}[10t^2 e^{-t}] = 10 \cdot [2t e^{-t} + t^2 (-e^{-t})] \].Simplify it to:\[ C'(t) = 10e^{-t}(2t - t^2) \].
04

Find the Maximum Concentration

Set \(C'(t) = 0\) to find critical points:\[ 10e^{-t}(2t - t^2) = 0 \].Since \(e^{-t} eq 0\), solve \(2t - t^2 = 0\):\[ t(t - 2) = 0 \Rightarrow t = 0, 2 \].Evaluate \(C(t)\) at critical points and boundaries to find:- Maximum concentration occurs at \(t = 2\) with \(C(2) \approx 5.41\).
05

Interpret the Derivative

The derivative \(C'(t) = 10e^{-t}(2t - t^2)\) indicates the rate at which the concentration changes over time. A positive \(C'(t)\) means the concentration is increasing, while a negative \(C'(t)\) means it is decreasing. The point where \(C'(t) = 0\) corresponds to where the maximum concentration occurs.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus focused on determining how a function changes at any given point. When you differentiate a function, you're essentially finding a new function that gives the slope or rate of change of the original function. In the context of our problem, the concentration of medication over time is given by the function \( C(t) = 10 t^2 e^{-t} \). By differentiating this function, we are finding how fast or slow the concentration changes over time.

To differentiate \( C(t) \), we apply the product and chain rules of differentiation. Specifically, we have the product of two functions: \( 10t^2 \) and \( e^{-t} \). The derivative is computed as:
  • \( \frac{d}{dt}[10t^2]e^{-t} + 10t^2 \frac{d}{dt}[e^{-t}] \) which gives us \( 2t \, 10e^{-t} \) and \( -t^2 \, 10e^{-t} \), respectively.
  • Combining results in: \( C'(t) = 10e^{-t}(2t - t^2) \).
The derivative \( C'(t) \) is then used to analyze how the concentration behaves at various points in time, showing both increasing and decreasing trends.
Graph Sketching
Graph sketching involves plotting the main features of a function to understand its behavior visually. For functions like \( C(t) = 10 t^2 e^{-t} \), the graph displays how the concentration of the medication changes over time.

To sketch the graph, you plot specific points that were already calculated:
  • \((0,0)\): The concentration starts at zero.
  • \((1,3.68)\): Concentration increases.
  • \((2,5.41)\): The concentration peaks.
  • \((3,4.48)\): Begins to decrease.
  • \((10,0.045)\): Close to zero reflects the exponential decay.
By connecting these points smoothly, the graph starts from the origin, rises to the maximum at \(t = 2\), and then gradually decreases, illustrating the medication being absorbed and then reducing as it's metabolized. Graph sketching is a helpful tool to visually interpret how functions like these behave over time.
Exponential Functions
Exponential functions are characterized by a constant raised to the power of a variable. They have unique properties, including growth or decay, depending on whether the exponent is positive or negative. In the function \( C(t) = 10 t^2 e^{-t} \), the \( e^{-t} \) component represents exponential decay, where the base \( e \) is Euler's number (approximately 2.718).

Exponential decay implies that as \( t \) increases, \( e^{-t} \) decreases rapidly towards zero. This property is why, in the context of our exercise, the concentration starts decreasing after a certain point (after peaking). The initial multiplier \( 10t^2 \) influences how quickly the concentration increases before the effect of decay becomes dominant.

Understanding exponential functions helps in problems involving continuous growth or decay, like population models, radioactive decay, and in this case, concentration levels in a biological system.

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