Problem 63
Question
Solve each problem. W(t)$ represents the gallons of water in a tank after I minutes. Complete the following. (a) Find the initial amount of water in the tank. (b) Find the amount of water in the rank after 12 minutes. (c) Is the rate of change in the amount of water in the rank constant? Explain. (d) Find the rate of change in the amount of water at 12 minutes. $$W(t)=500-t^{2}, \quad \text { for } 0 \leq t \leq 20$$
Step-by-Step Solution
Verified Answer
(a) 500 gallons. (b) 356 gallons. (c) No, the rate is not constant. (d) -24 gallons/min.
1Step 1: Find Initial Water Amount
To find the initial amount of water in the tank, evaluate the function \(W(t)\) at \(t = 0\). This represents the water present at the beginning. Substitute \(t = 0\) into the equation \(W(t) = 500 - t^2\). Result:\[W(0) = 500 - 0^2 = 500\] Thus, the initial amount of water is 500 gallons.
2Step 2: Water Amount After 12 Minutes
Next, we calculate the amount of water after 12 minutes by substituting \(t = 12\) into the equation. \[W(12) = 500 - (12)^2 = 500 - 144 = 356\] Therefore, the water amount after 12 minutes is 356 gallons.
3Step 3: Determine if Rate of Change is Constant
To know if the rate of change is constant, we examine if the function \(W(t) = 500 - t^2\) is linear. A linear function has a constant slope. Since \(W(t)\) is a quadratic function (\(-t^2\) term), it is not linear, indicating the rate of change is not constant.
4Step 4: Rate of Change at 12 Minutes
To find the instantaneous rate of change at \(t = 12\), compute the derivative \(W'(t)\) and evaluate at \(t = 12\). First, find the derivative:\[W'(t) = \frac{d}{dt}(500 - t^2) = -2t\]Now, substitute \(t = 12\):\[W'(12) = -2(12) = -24\]The rate of change in the amount of water at 12 minutes is \(-24\) gallons per minute.
Key Concepts
FunctionsQuadratic EquationsRate of ChangeDerivatives
Functions
When we talk about functions in mathematics, we're dealing with a special relationship. It's a relationship between a set of inputs and a set of outputs. For every input, there is exactly one output. This concept is crucial for understanding how the world works mathematically.
Let's consider the function in the exercise, represented by \(W(t)\). This function gives us the amount of water in a tank, with \(t\) being the time in minutes. So, every time we plug a different \(t\) into \(W(t)\), we get a different amount of water in the tank. It's like a machine that takes time as input and gives water amount as output.
Let's consider the function in the exercise, represented by \(W(t)\). This function gives us the amount of water in a tank, with \(t\) being the time in minutes. So, every time we plug a different \(t\) into \(W(t)\), we get a different amount of water in the tank. It's like a machine that takes time as input and gives water amount as output.
- Input (Independent Variable): Time \(t\) in minutes
- Output (Dependent Variable): Water in gallons \(W(t)\)
Quadratic Equations
A quadratic equation is any equation that can be rearranged into the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(aeq 0\). These equations feature a term with \(x^2\), making them non-linear and parabola-shaped on a graph.
In our example function, \(W(t) = 500 - t^2\), we actually see a quadratic equation at play. Here, the quadratic part is \(-t^2\). This tells us that the function will produce a parabolic shape when graphed. Specifically, the parabola will be upside down (opening downwards) because of the negative sign.
Quadratic equations like this often model real-world processes that involve acceleration or deceleration, like the flow of water. In this exercise, the \(t^2\) term impacts how the water quantity changes over time. Understanding these curves can help us predict when resources run out or fill up completely.
In our example function, \(W(t) = 500 - t^2\), we actually see a quadratic equation at play. Here, the quadratic part is \(-t^2\). This tells us that the function will produce a parabolic shape when graphed. Specifically, the parabola will be upside down (opening downwards) because of the negative sign.
Quadratic equations like this often model real-world processes that involve acceleration or deceleration, like the flow of water. In this exercise, the \(t^2\) term impacts how the water quantity changes over time. Understanding these curves can help us predict when resources run out or fill up completely.
Rate of Change
The rate of change is about understanding how one quantity changes in relation to another. In simple terms, it helps us understand how fast or slow something happens. When you look at how water level changes over time, you are examining a rate of change.
In this exercise, the question is whether the rate of change of water in the tank is constant. The function \(W(t) = 500 - t^2\) is not linear, it is quadratic. This means its rate of change is not constant over time. For a linear relationship like \(y = mx + b\), the rate of change (or slope) \(m\) is the same for all points. But in quadratic functions, this rate varies.
Why is this important? Because it tells us that the amount of water decreases at different rates over time. It's decreasing faster as time goes on. Recognizing these changing rates helps us anticipate future behavior of systems, whether it's how quickly a tank empties or how fast a vehicle accelerates.
In this exercise, the question is whether the rate of change of water in the tank is constant. The function \(W(t) = 500 - t^2\) is not linear, it is quadratic. This means its rate of change is not constant over time. For a linear relationship like \(y = mx + b\), the rate of change (or slope) \(m\) is the same for all points. But in quadratic functions, this rate varies.
Why is this important? Because it tells us that the amount of water decreases at different rates over time. It's decreasing faster as time goes on. Recognizing these changing rates helps us anticipate future behavior of systems, whether it's how quickly a tank empties or how fast a vehicle accelerates.
Derivatives
Derivatives are a key part of calculus, providing a way to measure how a function changes as its input changes. This is like a mathematical microscope: you can see exactly how the rate of change behaves at any point.
For this exercise, when you want to find the rate of change of water at exactly 12 minutes, you use the derivative. The derivative of our function \(W(t) = 500 - t^2\) is \(W'(t) = -2t\). What this tells us is the slope of the tangent to the curve at any point \(t\).
For this exercise, when you want to find the rate of change of water at exactly 12 minutes, you use the derivative. The derivative of our function \(W(t) = 500 - t^2\) is \(W'(t) = -2t\). What this tells us is the slope of the tangent to the curve at any point \(t\).
- Instantaneous Rate of Change: Evaluate at specific points (e.g., \(t = 12\))
- Slope of Tangent: Tells us how \(W(t)\) is changing at that moment
Other exercises in this chapter
Problem 62
Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. \(\lim _{x \rightarrow 0} \frac{2 x}{\tan x}\)
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Find two functions \(f(x)\) and \(g(x)\) with the given properties. $$\lim _{x \rightarrow x} f(x)=\infty, \lim _{x \rightarrow x} g(x)=\infty, \text { and } \l
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Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. \(\lim _{x \rightarrow 0}(x \csc x)\)
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