- In most of our applications of the derivative so far,
- we have been interested in the instantaneous rate at which one variable,
- say y, changes with respect to another, say x,
- leading us to compute and interpret \frac{dy}{dx}.
- We next consider situations where several variable quantities are related,
- but where each quantity is implicitly a function of time,
- which will be represented by the variable t.
- Through understanding how the quantities themselves are related,
- we will be able to determine how their respective rates of change with respect to time are related.
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- For example,
- suppose that air is being pumped into a spherical balloon so that its volume increases at a constant rate of 20 cubic inches per second.
- Since the balloon's volume and radius are related,
- by knowing how fast the volume is changing,
- we ought to be able to discover how fast the radius is changing.
- We are interested in questions such as:
- can we determine how fast the radius of the balloon is increasing at the moment the balloon's diameter is 12 inches?
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- Related Rates Problems
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- In problems where two or more quantities can be related to one another,
- and all of the variables involved are implicitly functions of time,
- t,
- we are often interested in how their rates are related;
- we call these related rates problems.
- related rates
- Once we have an equation establishing the relationship among the variables,
- we differentiate implicitly with respect to time to find connections among the rates of change.
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- Sand is being dumped by a conveyor belt onto a pile so that the sand forms a right circular cone,
- as pictured in Figure. How are the instantaneous rates of change of the sand's volume, height, and radius related to one another?
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A conical pile of sand.
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- As sand falls from the conveyor belt,
- several features of the sand pile will change:
- the volume of the pile will grow,
- the height will increase, and the radius will get bigger, too.
- All of these quantities are related to one another,
- and the rate at which each is changing is related to the rate at which sand falls from the conveyor.
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- We begin by identifying which variables are changing and how they are related.
- In this situation,
- we observe that the radius and height of the pile are related to its volume by the standard equation for the volume of a cone,
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- V = \frac{1}{3} \pi r^2 h
- .
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- Viewing each of V, r,
- and h as functions of t,
- we differentiate implicitly to arrive at an equation that relates their respective rates of change.
- Taking the derivative of each side of the equation with respect to t, we find
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- \frac{d}{dt}[V] = \frac{d}{dt}\left[\frac{1}{3} \pi r^2 h\right]
- .
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- On the left,
- \frac{d}{dt}[V] is simply \frac{dV}{dt}.
- On the right, the situation is more complicated,
- as both r and h are implicit functions of t.
- Hence we need the product and chain rules.
- We find that
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- \frac{dV}{dt} \amp= \frac{d}{dt}\left[\frac{1}{3} \pi r^2 h\right]
- \amp= \frac{1}{3} \pi r^2 \frac{d}{dt}[h] + \frac{1}{3} \pi h \frac{d}{dt}[r^2]
- \amp= \frac{1}{3} \pi r^2 \frac{dh}{dt} + \frac{1}{3} \pi h 2r \frac{dr}{dt}
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- (Note particularly how we are using ideas from Section on implicit differentiation.
- There we found that when y is an implicit function of x,
- \frac{d}{dx}[y^2] = 2y \frac{dy}{dx}.
- The same ideas are applied here when we compute \frac{d}{dt}[r^2] = 2r \frac{dr}{dt}.)
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- The equation
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- \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} + \frac{2}{3} \pi rh \frac{dr}{dt}
- ,
- relates the rates of change of V, h,
- and r.
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- If we are given sufficient additional information,
- we may then find the value of one or more of these rates of change at a specific point in time.
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- In the setting of Example,
- suppose we also know the following: (a) sand falls from the conveyor in such a way that the height of the pile is always half the radius,
- and (b) sand falls from the conveyor belt at a constant rate of 10 cubic feet per minute.
- How fast is the height of the sandpile changing at the moment the radius is 4 feet?
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- The information that the height is always half the radius tells us that for all values of t,
- h = \frac{1}{2}r.
- Differentiating with respect to t,
- it follows that \frac{dh}{dt} = \frac{1}{2} \frac{dr}{dt}.
- These relationships enable us to relate
- \frac{dV}{dt} to just one of r or h.
- Substituting the expressions involving r and
- \frac{dr}{dt} for h and \frac{dh}{dt},
- we now have that
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- \frac{dV}{dt} = \frac{1}{3} \pi r^2 \cdot \frac{1}{2} \frac{dr}{dt} + \frac{2}{3} \pi r \cdot \frac{1}{2}r \cdot \frac{dr}{dt}
- .
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- Since sand falls from the conveyor at the constant rate of 10 cubic feet per minute,
- the value of \frac{dV}{dt},
- the rate at which the volume of the sand pile changes,
- is \frac{dV}{dt} = 10 ft^3/min.
- We are interested in how fast the height of the pile is changing at the instant when r = 4, so
- we substitute r = 4 and
- \frac{dV}{dt} = 10 into Equation, to find
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- 10 = \frac{1}{3} \pi 4^2 \cdot \frac{1}{2} \left. \frac{dr}{dt} \right|_{r=4} + \frac{2}{3} \pi 4 \cdot \frac{1}{2}4 \cdot \left. \frac{dr}{dt} \right|_{r=4} = \frac{8}{3}\pi \left. \frac{dr}{dt} \right|_{r=4} + \frac{16}{3} \pi \left. \frac{dr}{dt} \right|_{r=4}
- .
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- Only the value of \left. \frac{dr}{dt} \right|_{r=4} remains unknown.
- We combine like terms on the right side of the equation above to get
- 10 = 8 \pi \left. \frac{dr}{dt} \right|_{r=4},
- and solve for \left. \frac{dr}{dt} \right|_{r=4} to find
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- \left. \frac{dr}{dt} \right|_{r=4} = \frac{10}{8\pi} \approx 0.39789
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- feet per minute.
- Because we were interested in how fast the height of the pile was changing at this instant,
- we want to know \frac{dh}{dt} when r = 4.
- Since \frac{dh}{dt} = \frac{1}{2} \frac{dr}{dt} for all values of t,
- it follows
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- \left. \frac{dh}{dt} \right|_{r=4} = \frac{5}{8\pi} \approx 0.19894 \ \text{ft/min}
- .
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- Observe the difference between the notations
- \frac{dr}{dt} and \left. \frac{dr}{dt} \right|_{r=4}.
- The former represents the rate of change of r with respect to t at an arbitrary value of t,
- while the latter is the rate of change of r with respect to t at a particular moment,
- the moment when r = 4.
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- Had we known that h = \frac{1}{2}r at the beginning of Example,
- we could have immediately simplified our work by writing V solely in terms of r to have
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- V = \frac{1}{3} \pi r^2 \left(\frac{1}{2}r\right) = \frac{1}{6} \pi r^3
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- From this last equation, differentiating with respect to t implies
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- \frac{dV}{dt} = \frac{1}{2} \pi r^2 \frac{dr}{dt}
- ,
- from which the same conclusions can be made.
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- Our work with the sandpile problem above is similar in many ways to our approach in Preview Activity,
- and these steps are typical of most related rates problems.
- In certain ways,
- they also resemble work we do in applied optimization problems,
- and here we summarize the main approach for consideration in subsequent problems.
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- Draw two or three possible figures that clearly represent the situation.
- Identify the quantities in the problem that are changing and choose clearly defined variable names for them.
- Identify the quantities or relationships that are not changing.
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- Determine all rates of change that are known or given and identify the rate(s) of change to be found.
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- Find an equation that relates the variables whose rates of change are known to those variables whose rates of change are to be found.
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- Differentiate implicitly with respect to t to relate the rates of change of the involved quantities.
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- Evaluate the derivatives and variables at the information relevant to the instant at which a certain rate of change is sought.
- Use proper notation to identify when a derivative is being evaluated at a particular instant,
- such as \left. \frac{dr}{dt} \right|_{r=4}.
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- When identifying variables and drawing pictures,
- it is important to think about the dynamic ways in which the quantities change.
- Usually a sequence of several pictures is helpful;
- for some pictures that can be easily modified as interactive built in Geogebra,
- see the following links,
- We again refer to the work of Prof.Marc Renault,
- found at
- gvsu.edu/s/5p.
- which represent
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- how a circular oil slick's area grows as its radius increases;
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- how the location of the base of a ladder and its height along a wall change as the ladder slides;
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- how the water level changes in a conical tank as it fills with water at a constant rate
- (compare the setting in Activity);
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- how a skateboarder's shadow changes as he moves past a lamppost.
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- Drawing well-labeled diagrams (usually several of them) and envisioning how different parts of the figure change is a key part of understanding related rates problems and being successful at solving them.
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- Recognizing which geometric relationships are relevant in a given problem
- often simplifies the process of differentiation when we relate the rates.
- For instance,
- although the problem in Activity
- is about a conical tank,
- the most important fact is that there are two similar right triangles involved.
- In another setting,
- we might use the Pythagorean Theorem to relate the legs of the triangle.
- But in the conical tank,
- the fact that the water fills the tank in such a way that that the ratio of radius to depth is constant turns out to be the important relationship.
- In other situations where a changing angle is involved,
- trigonometric functions may provide the means to find relationships among various parts of the triangle.
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- In addition to finding instantaneous rates of change at specific instants,
- we can often make more general observations about how particular rates themselves will change over time.
- For instance,
- when a conical tank is filling with water at a constant rate,
- it seems obvious that the depth of the water should increase more slowly over time.
- Note how carefully we must phrase the relationship:
- we mean to say that while the depth,
- h, of the water is increasing,
- its rate of change, \frac{dh}{dt}, is decreasing
- (both as a function of t and as a function of h).
- We make this observation by solving the equation that relates the various rates for one particular rate,
- without substituting any particular values for known variables or rates.
- For instance,
- in the conical tank problem in Activity,
- we established that
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- \frac{dV}{dt} = \frac{1}{16} \pi h^2 \frac{dh}{dt}
- ,
- and hence
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- \frac{dh}{dt} = \frac{16}{\pi h^2} \frac{dV}{dt}
- .
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- Provided that \frac{dV}{dt} is constant,
- it is apparent that as h gets larger,
- \frac{dh}{dt} will get smaller
- but remain positive.
- Hence, the depth of the water is increasing at a decreasing rate.
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- In the first three activities of this section,
- we provided guided instruction to build a solution in a step by step way.
- For the closing activity and the following exercises,
- most of the detailed work is left to the reader.
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- Summary
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- When two or more related quantities are changing as implicit functions of time,
- their rates of change can be related by implicitly differentiating the equation that relates the quantities themselves.
- For instance,
- if the sides of a right triangle are all changing as functions of time,
- say having lengths x, y, and z,
- then these quantities are related by the Pythagorean Theorem:
- x^2 + y^2 = z^2.
- It follows by implicitly differentiating with respect to t that their rates are related by the equation
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- 2x \frac{dx}{dt} + 2y\frac{dy}{dt} = 2z \frac{dz}{dt}
- ,
- so that if we know the values of x, y,
- and z at a particular time,
- as well as two of the three rates,
- we can deduce the value of the third.
-