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Water is flowing into a tank; the depth, in feet, of the water at time \(t\) in hours is \(h(t) .\) Interpret, with units, the following statements. (a) \(\quad h(5)=3\) (b) \(h^{\prime}(5)=0.7\) (c) \(h^{-1}(5)=7\) (d) \(\quad\left(h^{-1}\right)^{\prime}(5)=1.2\)

Short Answer

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(a) At 5 hours, depth is 3 feet. (b) Depth increases 0.7 feet/hour at 5 hours. (c) Depth of 5 feet occurs at 7 hours. (d) For 5 feet, time changes 1.2 hours/foot.

Step by step solution

01

Interpret the function value h(5)=3

The equation \( h(5) = 3 \) tells us that at time \( t = 5 \) hours, the depth of the water in the tank is 3 feet. This means the function \( h(t) \) gives the water depth directly at given times.
02

Interpret the derivative h'(5)=0.7

The derivative \( h'(5) = 0.7 \) indicates the rate at which the water level is changing at \( t = 5 \) hours. Specifically, it means that at this time, the water level is increasing at a rate of 0.7 feet per hour.
03

Interpret the inverse function h鈦宦(5)=7

The statement \( h^{-1}(5) = 7 \) means when the water depth is 5 feet, it corresponds to \( t = 7 \) hours. The function \( h^{-1}(t) \) provides the time at which a particular water depth is reached.
04

Interpret the derivative of the inverse (h鈦宦)'(5)=1.2

The expression \( \left(h^{-1}\right)'(5) = 1.2 \) implies that when the water depth changes from 5 feet, the rate of change of the time required to reach this depth is 1.2 hours per foot. This tells us about the sensitivity of time with respect to changes in depth at 5 feet.

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Key Concepts

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

Derivatives
Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. In this context, the derivative of the function, noted as \( h'(t) \), indicates how quickly the water level in the tank changes over time. For example, the derivative \( h'(5) = 0.7 \) tells us that at 5 hours, the water depth is increasing by 0.7 feet every hour.
This rate of change is instantaneous, meaning it reflects the slope of the tangent line to the curve of the function at the specific point \( t=5 \). Understanding derivatives helps us predict how systems behave over very short intervals, giving insight into patterns of increase or decrease in the context of real-world phenomena like water levels in a tank. This concept is crucial in analyzing trends and dynamics in various fields such as physics, engineering, and economics.
Inverse Functions
Inverse functions are used to reverse the roles of inputs and outputs for the original function. For a function \( h(t) \), its inverse \( h^{-1}(t) \) swaps the dependent and independent variables, allowing us to find the input that corresponds to a given output. In practical terms, if \( h^{-1}(5) = 7 \), it means that a water depth of 5 feet is reached at 7 hours.
Understanding inverse functions is essential when we are dealing with scenarios where outputs are given, and we need to determine the corresponding inputs. This is common in many real-world problems, particularly those involving time and quantity, such as this tank example, where we might know the desired depth and want to know when it will occur. This knowledge helps us answer questions of 'when did this happen?' or 'how long until a certain event?'.
Rate of Change
The rate of change is a measure of how much a quantity increases or decreases over time. In this example, the derivative \( h'(t) \) represents the rate at which the water level changes as time passes. Specifically, \( h'(5) = 0.7 \) indicates that the water level is rising 0.7 feet per hour at \( t = 5 \).
This notion is essential because it simplifies understanding how systems behave, especially in situations involving growth, decline, speed, or other transitions over time. Rates of change can be constant or variable, and analyzing them can help identify trends, forecast future behavior, and make informed decisions in various scientific and engineering fields. It forms the backbone of interpreting how changes affect the system being studied.
Function Interpretation
Interpreting functions involves understanding what each part of a function and its expressions mean in a real-world context. When given \( h(t) \) such as in this example, \( h(5) = 3 \) simply states that at 5 hours, the depth of water is 3 feet. Similarly, \( h^{-1}(5) \) and its derivative reveal more about timing and change concerning depth.
Each function or expression tells a specific story. By learning to interpret functions correctly, we gain a clearer picture of the situation at hand, allowing for effective analysis and problem-solving. This ability to decode various expressions highlights not only relationships between variables but also how one variable influences another, which is vital in various branches of science and mathematics.

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Most popular questions from this chapter

When you breathe, a muscle (called the diaphragm) reduces the pressure around your lungs and they expand to fill with air. The table shows the volume of a lung as a function of the reduction in pressure from the diaphragm. Pulmonologists (lung doctors) define the compliance of the lung as the derivative of this function. \(^{9}\) (a) What are the units of compliance? (b) Estimate the maximum compliance of the lung. (c) Explain why the compliance gets small when the lung is nearly full (around 1 liter). $$\begin{array}{|c|c|}\hline \begin{array}{c}\text { Pressure reduction } \\\\\text { (cm of water) } \end{array} & \begin{array}{c}\text { Volume } \\\\\text { (liters) }\end{array} \\\\\hline 0 & 0.20 \\ \hline 5 & 0.29 \\\\\hline 10 & 0.49 \\\\\hline 15 & 0.70 \\\\\hline 20 & 0.86 \\\\\hline 25 & 0.95 \\ \hline 30 & 1.00 \\\\\hline\end{array}$$

A ball is tossed into the air from a bridge, and its height, \(y\) (in feet), above the ground \(t\) seconds after it is thrown is given by $$ y=f(t)=-16 t^{2}+50 t+36 $$ (a) How high above the ground is the bridge? (b) What is the average velocity of the ball for the first second? (c) Approximate the velocity of the ball at \(t=1 \mathrm{sec}\) ond. (d) Graph \(f,\) and determine the maximum height the ball reaches. What is the velocity at the time the ball is at the peak? (e) Use the graph to decide at what time, \(t,\) the ball reaches its maximum height.

Find the equation of the line tangent to the function at the given point.$$f(x)=x\( at \)x=20$$

Chlorofluorocarbons (CFCs) were used as propellants in spray cans until their build up in the atmosphere started destroying the ozone, which protects us from ultraviolet rays. since the 1987 Montreal Protocol (an agreement to curb CFCs), the CFCs in the atmosphere above the US have been reduced from a high of 3200 parts per trillion (ppt) in 1994 to 2750 ppt in \(2010 .^{15}\) The reduction has been approximately linear. Let \(C(t)\) be the concentration of CFCs in ppt in year \(t\) (a) Find \(C(1994)\) and \(C(2010)\) (b) Estimate \(C^{\prime}(1994)\) and \(C^{\prime}(2010)\) (c) Assuming \(C(t)\) is linear, find a formula for \(C(t)\) (d) When is \(C(t)\) expected to reach 1850 ppt, the level before CFCs were introduced? (e) If you were told that in the future, \(C(t)\) would not be exactly linear, and that \(C^{\prime \prime}(t)>0,\) would your answer to part (d) be too early or too late?

Explain what is wrong with the statement. If \(f\) is not differentiable at a point then it is not continuous at that point.

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