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Chapter 3: Problem 28

The number of liters of water in a tank \(t\) minutes after the tank has started to drain is \(Q(t)=200(30-t)^{2}.\) How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min?

Short Answer

Expert verified
Water runs out at 8000 L/min at 10 min; average flow rate is -10000 L/min.

Step by step solution

01

Understanding the Formula

Given the function \( Q(t) = 200(30-t)^2 \), we understand that \( Q(t) \) represents the amount of water in the tank at any time \( t \). Our task includes finding the instantaneous rate of change at \( t = 10 \) and the average rate over the first 10 minutes.
02

Finding the Instantaneous Rate of Change

The instantaneous rate of change of \( Q(t) \) at any time \( t \) is given by its derivative, \( Q'(t) \). First, compute the derivative: \( Q'(t) = 200 \cdot 2(30-t)(-1) = -400(30-t) \). At \( t = 10 \), this becomes \( Q'(10) = -400(30-10) = -400 \times 20 = -8000 \). Thus, the water is running out at 8000 liters per minute when \( t = 10 \) minutes.
03

Calculating the Average Rate of Change

The average rate of change over the first 10 minutes is found using the formula: \( \frac{Q(10) - Q(0)}{10 - 0} \). First, calculate these quantities: \( Q(0) = 200(30)^2 = 200 \times 900 = 180000 \) and \( Q(10) = 200(20)^2 = 200 \times 400 = 80000 \). Now, \( \text{Average Rate} = \frac{80000 - 180000}{10} = \frac{-100000}{10} = -10000 \). Hence, the average rate is -10000 liters per minute.

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Key Concepts

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

Average Rate of Change
The average rate of change is an essential concept in understanding how a quantity changes over an interval of time. In the given problem, we are exploring how the water level in a tank changes from the start up to 10 minutes.

To calculate the average rate:
  • First, evaluate the water level at the beginning: \( Q(0) = 200(30)^2 = 180000 \) liters.
  • Then, evaluate the water at 10 minutes: \( Q(10) = 200(20)^2 = 80000 \) liters.
  • The change in water between these times is \( Q(10) - Q(0) = 80000 - 180000 = -100000 \) liters.
Divide the water change by the time interval (10 minutes) to find the average rate:
  • Average Rate = \( \frac{-100000}{10} = -10000 \) liters per minute.
This negative value indicates that the water level is decreasing over time.
Derivative Calculation
The derivative is a key tool in calculus, used to find the instantaneous rate of change of a function. In this problem, the derivative helps to determine how quickly the water is leaving the tank at exactly 10 minutes.

For the function \( Q(t) = 200(30-t)^2 \), the derivative \( Q'(t) \) gives us the rate of change. Calculate it as follows:
  • Differentiate \( Q(t) \) using the chain rule: \( Q'(t) = 200 \cdot 2(30-t)(-1) \).
  • Simplify to: \( Q'(t) = -400(30-t) \).
At \( t = 10 \), substitute to find the rate of change:
  • \( Q'(10) = -400(20) = -8000 \) liters per minute.
This derivative value tells us that at 10 minutes, the water drains at 8000 liters per minute, showing a rapid decrease.
Functions and Modeling
Functions form the backbone of mathematical modeling, helping us represent real-world scenarios. In this exercise, the function \( Q(t) = 200(30-t)^2 \) provides a model of the draining tank.

Here's why functions are crucial:
  • They describe relationships between variables, such as time and water volume here.
  • They allow predictions and understanding of behavior over time.
In this scenario:
  • The function represents the quantity of water as a function of time.
  • Analyzing the function's derivative and average rate helps us understand both the overall trend and specific moments in time.
This kind of modeling is indispensable in fields ranging from engineering to environmental science.
Calculus Applications
Calculus provides powerful tools to solve a variety of real-world problems, including those involving rates of change and accumulations.

Applications of calculus in this problem include:
  • Using derivatives to find instantaneous rates of change, which help us understand quick snapshots of behavior, such as the rapid water draining rate at a specific time.
  • Applying average rate of change calculations to see the general trend of water outflow over a period, which might help in planning and managing resources.
Understanding these applications:
  • Helps in systems where rates of change need monitoring or control.
  • Is crucial for designing systems with optimal performance and efficiency.
Overall, calculus translates complex dynamical systems into understandable models and guides efficient decision-making.

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Most popular questions from this chapter

A particle moves along the \(x\) -axis with velocity \(d x / d t=f(x) .\) Show that the particle's acceleration is \(f(x) f^{\prime}(x)\)

Find the tangent to \(y=\sqrt{x^{2}-x+7}\) at \(x=2\)

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? \(s=60 t-4.9 t^{2}, \quad 0 \leq t \leq 12.5\) (a heavy object fired straight up from Earth's surface at \(60 \mathrm{m} / \mathrm{s}\) )

Suppose that a piston is moving straight up and down and that its position at time \(t\) s is $$s=A \cos (2 \pi b t)$$ with \(A\) and \(b\) positive. The value of \(A\) is the amplitude of the motion, and \(b\) is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston's velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.)

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x^{5}+y^{3} x+y x^{2}+y^{4}=4, \quad P(1,1)$$
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