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Chapter 10: Problem 14

Suppose that the temperature on a metal plate is given by the function \(T\) with $$T(x, y)=100-\left(x^{2}+4 y^{2}\right)$$ where the temperature is measured in degrees Fahrenheit and \(x\) and \(y\) are each measured in feet. Now suppose that an ant is walking on the metal plate in such a way that it walks in a straight line from the point (1,4) to the point (5,6) . a. Find parametric equations \((x(t), y(t))\) for the ant's coordinates as it walks the line from (1,4) to (5,6) b. What can you say about \(\frac{d x}{d t}\) and \(\frac{d y}{d t}\) for every value of \(t ?\) c. Determine the instantaneous rate of change in temperature with respect to \(t\) that the ant is experiencing at the moment it is halfway from (1,4) to \((5,6),\) using your parametric equations for \(x\) and \(y\). Include units on your answer.

Short Answer

Expert verified
The parametric equations for the ant's coordinates are \(x(t) = 1 + 4t\) and \(y(t) = 4 + 2t\). The rates of change of x and y coordinates with respect to time t are constant: \(\frac{dx}{dt} = 4\) and \(\frac{dy}{dt} = 2\). The instantaneous rate of change in temperature with respect to time t at the midpoint of the ant's path is -104 掳F/s.

Step by step solution

01

Find the parametric equations (x(t), y(t))

To find the parametric equations for the ant's coordinates, we will express the linear path between points (1,4) and (5,6) in terms of the variable t which represents the time. We can represent the x coordinate as: \[x(t) = x_1 + (x_2 - x_1)t\] Similarly, we can represent the y coordinate as: \[y(t) = y_1 + (y_2 - y_1)t\] In this case, our starting point is (1,4) and our ending point is (5,6). Therefore: \[x(t) = 1 + (5 - 1)t = 1 + 4t\] \[y(t) = 4 + (6 - 4)t = 4 + 2t\]
02

Analyze the rates of change with respect to time t

To find the rates of change for x and y coordinates with respect to time t, we will differentiate the parametric equations obtained in step 1: \[\frac{dx}{dt} = \frac{d(1 + 4t)}{dt} = 4\] \[\frac{dy}{dt} = \frac{d(4 + 2t)}{dt} = 2\] These rates are constant and do not depend on the value of t. This means that the ant's coordinates are changing at a constant rate with respect to time.
03

Determine the instantaneous rate of change in temperature with respect to time t at the midpoint

To find the instantaneous rate of change in temperature with respect to time, we will use the chain rule and the given temperature function: \[T(x, y) = 100 - (x^2 + 4y^2)\] First, let's find the midpoint of the ant's path which is halfway between points (1,4) and (5,6): Midpoint = \(\left(\frac{1+5}{2}, \frac{4+6}{2}\right) = (3, 5)\) At the midpoint, we will have: \[x(t) = 1 + 4t = 3 \Rightarrow t = \frac{1}{2}\] \[y(t) = 4 + 2t = 5 \Rightarrow t = \frac{1}{2}\] Now, we can apply the chain rule: \[\frac{dT}{dt} = \frac{dT}{dx} \cdot \frac{dx}{dt} + \frac{dT}{dy} \cdot \frac{dy}{dt}\] We need to find the derivatives of the temperature function with respect to x and y: \[\frac{dT}{dx} = -2x\] \[\frac{dT}{dy} = -8y\] At the midpoint, \(x = 3\) and \(y = 5\): \[\frac{dT}{dx} = -2(3) = -6\] \[\frac{dT}{dy} = -8(5) = -40\] Now, using the chain rule: \[\frac{dT}{dt} = (-6)(4) + (-40)(2) = -104\] The ant at the midpoint experiences a rate of change in temperature of -104 掳F/s.

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Key Concepts

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

Parametric Equations
In multivariable calculus, parametric equations are a powerful tool used to represent paths or curves in space by expressing each coordinate as a function of a single parameter, often labeled as \( t \). This comes in handy when describing objects that move or change over time.
To understand how parametric equations work, consider the ant's path on the metal plate moving from point (1,4) to (5,6). We express the \( x \)-coordinate and \( y \)-coordinate in terms of \( t \):
  • \( x(t) = 1 + 4t \)
  • \( y(t) = 4 + 2t \)
Here, \( t \) represents time, and as it increases from 0 to 1, the ant traces a linear path on the plate from the starting position to the ending position. Such a representation not only simplifies the understanding of motion but also allows us to manage more complex calculations later on.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. This comes into play when dealing with multivariable functions where one needs to understand how changes in input variables affect the output. It's especially useful for expressing rates of change.
For the problem with the ant on the plate, we have a temperature function \( T(x, y) \). To find how temperature changes with respect to time \( t \), while the ant moves, we utilize the chain rule:
\[ \frac{dT}{dt} = \frac{dT}{dx} \cdot \frac{dx}{dt} + \frac{dT}{dy} \cdot \frac{dy}{dt} \]Where:
  • \( \frac{dT}{dx} \) and \( \frac{dT}{dy} \) are partial derivatives of \( T \) with respect to \( x \) and \( y \), respectively.
  • \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are derivatives of the parametric equations.
This approach allows us to directly link the ant鈥檚 path (via \( t \)) to changes in temperature along that path.
Rate of Change
The rate of change in mathematics refers to how a quantity, such as temperature, varies with respect to another quantity, like time. It's an important concept in many areas, including physics and engineering.
In our exercise, as the ant walks along the plate, we want to know how fast the temperature changes due to its movement. The rate of change is calculated using derivatives. In particular, the instantaneous rate of change of temperature \( \left( \frac{dT}{dt} \right) \) at the midpoint of the ant鈥檚 path gives us precise information about how the temperature is decreasing or increasing at that instance.
  • The rate was found to be \(-104 \text{ } ^\circ F/s\), indicating a rapid decrease in temperature as the ant passes the midpoint.
Understanding this rate of change is crucial for evaluating dynamic systems where temperature fluctuations are significant.
Temperature Gradient
The temperature gradient is a vector that describes the rate and direction of temperature change in space. In mathematical terms, it is the gradient of the temperature function \( T(x, y) \), represented as \( abla T \).
For the given function \( T(x, y)=100-(x^2+4y^2) \), the gradient is calculated as follows:
  • \( \frac{dT}{dx} = -2x \)
  • \( \frac{dT}{dy} = -8y \)
At any point on the plate, these partial derivatives tell us how steeply the temperature changes in the \( x \) and \( y \) directions respectively. For example, at the midpoint \( (3, 5) \), these values were \(-6\) and \(-40\), respectively.
The temperature gradient gives insight into how heat might circulate over the plate and is key in simulations and predictions in thermal dynamics and engineering problems.

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Most popular questions from this chapter

For each of the following prompts, provide an example of a function of two variables with the desired properties (with justification), or explain why such a function does not exist. a. A function \(p\) that is defined at \((0,0),\) but \(\lim _{(x, y) \rightarrow(0,0)} p(x, y)\) does not exist. b. A function \(q\) that does not have a limit at \((0,0),\) but that has the same limiting value along any line \(y=m x\) as \(x \rightarrow 0\). c. A function \(r\) that is continuous at \((0,0),\) but \(\lim _{(x, y) \rightarrow(0,0)} r(x, y)\) does not exist. d. A function \(s\) such that \(\lim _{(x, x) \rightarrow(0,0)} s(x, x)=3\) and \(\lim _{(x, 2 x) \rightarrow(0,0)} s(x, 2 x)=6\) for which \(\lim _{(x, y) \rightarrow(0,0)} s(x, y)\) exists. e. A function \(t\) that is not defined at (1,1) but \(\lim _{(x, y) \rightarrow(1,1)} t(x, y)\) does exist.

The Heat Index, \(I,\) (measured in apparent degrees \(F)\) is a function of the actual temperature \(T\) outside (in degrees \(\mathrm{F}\) ) and the relative humidity \(H\) (measured as a percentage). A portion of the table which gives values for this function, \(I=I(T, H),\) is reproduced in Table \(10.2 .10 .\) $$\begin{array}{ccccc}\hline T \downarrow \backslash H \rightarrow & 70 & 75 & 80 & 85 \\ \hline 90 & 106 & 109 & 112 & 115 \\\\\hline 92 & 112 & 115 & 119 & 123 \\\\\hline 94 & 118 & 122 & 127 & 132 \\\\\hline 96 & 125 & 130 & 135 & 141 \\\\\hline\end{array}$$ a. State the limit definition of the value \(I_{T}(94,75)\). Then, estimate \(I_{T}(94,75),\) and write one complete sentence that carefully explains the meaning of this value, including its units. b. State the limit definition of the value \(I_{H}(94,75)\). Then, estimate \(I_{H}(94,75),\) and write one complete sentence that carefully explains the meaning of this value, including its units. c. Suppose you are given that \(I_{T}(92,80)=3.75\) and \(I_{H}(92,80)=0.8\). Estimate the values of \(I(91,80)\) and \(I(92,78)\). Explain how the partial derivatives are relevant to your thinking. d. On a certain day, at 1 p.m. the temperature is 92 degrees and the relative humidity is \(85 \%\). At 3 p.m., the temperature is 96 degrees and the relative humidity \(75 \% .\) What is the average rate of change of the heat index over this time period, and what are the units on your answer? Write a sentence to explain your thinking.

Consider the function $$f(x, y)=\left(18 x-x^{2}\right)\left(18 y-y^{2}\right)$$ Find and classify all critical points of the function. If there are more blanks than critical points, leave the remaining entries blank. \(f_{x}=\) _________. \(f_{y}=\) _________. \(f_{x x}=\) _________. \(f_{x y}=\) _________. \(f_{y y}=\) _________. There are several critical points to be listed. List them lexicograhically, that is in ascending order by x-coordinates, and for equal x-coordinates in ascending order by y-coordinates (e.g., (1,1),(2,-1),(2,3) is a correct order) The critical point with the smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest \(\mathrm{x}\) -coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest \(x\) -coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined)

Your monthly car payment in dollars is \(P=f\left(P_{0}, t, r\right),\) where \(\$ P_{0}\) is the amount you borrowed, t is the number of months it takes to pay off the loan, and r percent is the interest rate. (a) Is \(\partial P / \partial t positive or negative? \)\quad(\square\( positive \)\square\( negative) Suppose that your bank tells you that the magnitude of \)\partial P / \partial t is 15 . What are the units of this value? _________ (b) Is \(\partial P / \partial r positive or negative? \)\quad(\square\( positive \)\square$ negative) . What are the units of this value? __________.

In this section we argued that if \(f=f(x, y)\) is a function of two variables and if \(f_{x}\) and \(f_{y}\) both exist and are continuous in an open disk containing the point \(\left(x_{0}, y_{0}\right),\) then \(f\) is differentiable at \(\left(x_{0}, y_{0}\right) .\) This condition ensures that \(f\) is differentiable at \(\left(x_{0}, y_{0}\right),\) but it does not define what it means for \(f\) to be differentiable at \(\left(x_{0}, y_{0}\right) .\) In this exercise we explore the definition of differentiability of a function of two variables in more detail. Throughout, let \(g\) be the function defined by \(g(x, y)=\sqrt{|x y|}\) a. Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1] \times[-1,1] .\) Explain why \(g\) is not locally linear at (0,0) b. Show that both \(g_{x}(0,0)\) and \(g_{y}(0,0)\) exist. If \(g\) is locally linear at \((0,0),\) what must be the equation of the tangent plane \(L\) to \(g\) at (0,0)\(?\) c. Recall that if a function \(f=f(x)\) of a single variable is differentiable at \(x=x_{0},\) then $$f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$ exists. We saw in single variable calculus that the existence of \(f^{\prime}\left(x_{0}\right)\) means that the graph of \(f\) is locally linear at \(x=x_{0}\). In other words, the graph of \(f\) looks like its linearization \(L(x)=f\left(x_{0}\right)+\) \(f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\) for \(x\) close to \(x_{0} .\) That is, the values of \(f(x)\) can be closely approximated by \(L(x)\) as long as \(x\) is close to \(x_{0}\). We can measure how good the approximation of \(L(x)\) is to \(f(x)\) with the error function $$E(x)=L(x)-f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-f(x)$$ As \(x\) approaches \(x_{0}, E(x)\) approaches \(f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(0)-f\left(x_{0}\right)=0\), and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\) gets closer to \(x_{0} .\) Show that, even though \(g(x, y)=\sqrt{|x y|}\) is not locally linear at \((0,0),\) its error term $$ E(x, y)=L(x, y)-g(x, y) $$ at (0,0) has a limit of 0 as \((x, y)\) approaches \((0,0) .\) (Use the linearization you found in part (b).) This shows that just because an error term goes to 0 as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right),\) we cannot conclude that a function is locally linear at \(\left(x_{0}, y_{0}\right)\). d. As the previous part illustrates, having the error term go to 0 does not ensure that a function of two variables is locally linear. Instead, we need a notation of a relative error. To see how this works, let us return to the single variable case for a moment and consider \(f=f(x)\) as a function of one variable. If we let \(x=x_{0}+h,\) where \(|h|\) is the distance from \(x\) to \(x_{0}\), then the relative error in approximating \(f\left(x_{0}+h\right)\) with \(L\left(x_{0}+h\right)\) is $$\frac{E\left(x_{0}+h\right)}{h}$$ Show that, for a function \(f=f(x)\) of a single variable, the limit of the relative error is 0 as \(h\) approaches 0 . e. Even though the error term for a function of two variables might have a limit of 0 at a point, our example shows that the function may not be locally linear at that point. So we use the concept of relative error to define differentiability of a function of two variables. When we consider differentiability of a function \(f=f(x, y)\) at a point \(\left(x_{0}, y_{0}\right),\) then if \(x=x_{0}+h\) and \(y=y_{0}+k,\) the distance from \((x, y)\) to \(\left(x_{0}, y_{0}\right)\) is \(\sqrt{h^{2}+k^{2}}\)
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