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Active Calculus - Multivariable

Section 11.3 First-Order Partial Derivatives

Subsection 11.3.1 Introduction

The derivative plays a central role in single variable calculus because it provides important information about a function in terms of each of the representations of the function. Thinking graphically, the derivative at a point tells us the slope of the line tangent to the graph at the given point. Numerically, the derivative at a point also provides the instantaneous rate of change of the function with respect to a change in the independent variable. Algebraically, we can calculate the derivative as a function of the same input variable, which is useful as a tool to examine where the function is increasing or decreasing, as well as the concavity of the function’s graph.
In this chapter, we are investigating functions of two or more variables and we can still ask how fast a function of several variables is changing, but we have to be careful about what we mean. Thinking graphically again, we can try to measure how steep the graph of the function is in a particular direction. As we saw in Section 11.2, the idea of measuring the behavior of a two variable function in a particular direction and generalizing to other directions is a much deeper idea when you have multiple input variables. Alternatively, we may want to know how quickly a function’s output is changing in response to varying only one of the inputs. Over the next few sections, we will develop tools for addressing these questions and more. In the following preview activity, we will look at how we can apply our single variable calculus knowledge to measure the rate of change when we hold all but one input variable constant.

Preview Activity 11.3.1.
Suppose we take out an $18,000 car loan at interest rate \(r\) and we agree to pay off the loan in \(t\) years. The monthly payment, in dollars, is
\begin{equation*} M(r,t) = \frac{1500r}{1-\left(1+\frac{r}{12}\right)^{-12t}}. \end{equation*}
(a)
If you take a loan with an interest rate of \(3\%\) (so \(r = 0.03\)) and will pay it off in \(t=4\) years, what will your monthly payment be?
(b)
Suppose the interest rate on loans is fixed at \(3\%\text{.}\) Express \(M\) as a function \(f\) of \(t\) alone using \(r=0.03\text{.}\) That is, let \(f(t) = M(0.03, t)\text{.}\) Sketch the graph of \(f\) on Figure 11.3.1 and write a couple of sentences to explain the meaning of the function \(f\text{.}\)
Figure 11.3.1. A plot of \(f(t)=M(0.03,t)\)
(c)
Find the instantaneous rate of change \(f'(4)\) and state the units on this quantity. What information does \(f'(4)\) tell us about our car loan? What information does \(f'(4)\) tell us about the graph you sketched in (b)?
(d)
Suppose that we want to examine how the monthly payment will change if we plan to take 4 years to pay off the loan. Express \(M\) as a function of the interest rate \(r\) with a fixed loan term of \(t=4\text{.}\) That is, let \(g(r) = M(r, 4)\text{.}\) Give the function \(g(r)\) and sketch a graph of \(g\) on Figure 11.3.2. Write a couple of sentences to explain the meaning of the function \(g\text{.}\)
Figure 11.3.2. A plot of \(f(t)=M(0.03,t)\)
(e)
Find the instantaneous rate of change \(g'(0.03)\) and state the units on this quantity. Write a couple of sentences to explain what information does \(g'(0.03)\) tell us about our car loan. What information does \(g'(0.03)\) tell us about the graph you sketched in (d)?

Subsection 11.3.2 First-Order Partial Derivatives

In Section 11.1, we studied the behavior of a function of two or more variables by considering the traces of the function. By holding one of our input variables constant, we restricted our multivariable function to have only one input variable, which means we can use all of our calculus tools from the past several courses on these traces.
Recall that in one example, we considered the function \(f\) defined by
\begin{equation*} f(x,y) = \frac{x^2 \sin(2 y)}{32}, \end{equation*}
which measures the range, or horizontal distance, in feet, traveled by a projectile launched with an initial speed of \(x\) feet per second at an angle \(y\) radians to the horizontal. The graph of this function is given again in Figure 11.3.3 with traces plotted on the surface for \(x=150\) (in blue) and \(y=0.6\) (in red).
Figure 11.3.3. The distance surface given by \(z=f(x,y) = \frac{x^2 \sin(2y)}{g}\)
Moreover, if we fix the angle \(y = 0.6\text{,}\) we may view the trace \(f(x,0.6)\) as a function of \(x\) alone, as seen in Figure 11.3.4.
Figure 11.3.4. A plot of \(f(x,0.6)\)
Since the trace is a one-variable function, we may consider its derivative just as we did in single variable calculus. With \(y=0.6\text{,}\) we have
\begin{equation*} f(x,0.6) = \frac{\sin(1.2)}{32}x^2, \end{equation*}
and therefore
\begin{equation*} \frac{d}{dx}\Bigl[ f(x,0.6) \Bigr] = \frac{\sin(1.2)}{16}x. \end{equation*}
When \(x=150\text{,}\) this gives
\begin{equation*} \frac{d}{dx}\Bigr[ f(x,0.6) \Bigr]\restrict{x=150} = \frac{\sin(1.2)}{16}150 \approx 8.74~\mbox{feet per feet per second} , \end{equation*}
which gives the slope of the tangent line shown in Figure 11.3.4. Thinking of this derivative as an instantaneous rate of change implies that if we increase the initial speed of the projectile by one foot per second, we expect the horizontal distance traveled to increase by approximately 8.74 feet when we fix the launch angle at \(0.6\) radians. Note here that the sign of the derivative of the trace is positive, which is why there would be an increase to the range of the projectile if we increased the initial speed (above \(150 ft/s\)) while holding the launch angle constant at \(0.6\) radians.
By holding \(y\) fixed and differentiating with respect to \(x\text{,}\) we obtain the first-order partial derivative of \(f\) with respect to \(x\). Denoting this partial derivative as \(f_x\text{,}\) we have seen that
\begin{equation*} f_x(150, 0.6) = \frac{d}{dx} \Bigl[ f(x,0.6) \Bigr] \restrict{x=150} = \lim_{h\to 0}\frac{f(150+h, 0.6) - f(150, 0.6)}{h} \end{equation*}
More generally, we have
\begin{equation*} f_x(a,b) = \lim_{h \to 0} \frac{f(a+h, b)-f(a,b)}{h}, \end{equation*}
provided this limit exists.
In the same way, we may obtain a trace by setting, say, \(x=150\) as shown in Figure 11.3.5.
Figure 11.3.5. A plot of \(f(x,0.6)\)
This gives
\begin{equation*} f(150, y) = \frac{150^2}{32}\sin(2y), \end{equation*}
and therefore
\begin{equation*} \frac{d}{dy}[f(150,y)] = \frac{150^2}{16}\cos(2y). \end{equation*}
If we evaluate this quantity at \(y=0.6\text{,}\) we have
\begin{equation*} \frac{d}{dy}\Bigl[ f(150,y) \Bigr]\restrict{y=0.6} = \frac{150^2}{16}\cos(1.2) \approx 509.5 ~\mbox{feet per radian} \end{equation*}
Once again, the derivative gives the slope of the tangent line shown on the right in Figure 11.3.5. Thinking of the derivative as an instantaneous rate of change, we expect that the range of the projectile increases by 509.5 feet for every radian we increase the launch angle \(y\) if we keep the initial speed of the projectile constant at 150 feet per second.
By holding \(x\) fixed and differentiating with respect to \(y\text{,}\) we obtain the first-order partial derivative of \(f\) with respect to \(y\). As before, we denote this partial derivative as \(f_y\) and write
\begin{equation*} f_y(150, 0.6) = \frac{d}{dy} \Bigl[ f(150,y) \Bigr]\restrict{y=0.6} = \lim_{h \to 0}\frac{f(150, 0.6+h) - f(150, 0.6)}{h}. \end{equation*}
As with the partial derivative with respect to \(x\text{,}\) we may express this quantity more generally at an arbitrary point \((a,b)\text{.}\) To recap, we have now arrived at the formal definition of the first-order partial derivatives of a function of two variables.

Definition 11.3.6.

The first-order partial derivatives of \(f\) with respect to \(x\) and \(y\) at a point \((a,b)\) are (provided the limits exist), respectively,
\begin{align*} f_x(a,b) \amp = \frac{\partial f}{\partial x} \amp = \lim_{h \to 0} \frac{f(a+h,b)-f(a,b)}{h}\\ f_y(a,b) \amp = \frac{\partial f}{\partial y} \amp = \lim_{h \to 0} \frac{f(a,b+h)-f(a,b)}{h} \end{align*}

Activity 11.3.2.
In this activity, we will be examining the function \(f\) defined by
\begin{equation*} f(x,y) = \frac{x y^2}{x+1} \end{equation*}
at the point \((1,2)\text{.}\) Specifically, we will be looking how the traces of this function will be related to the values of the partial derivatives.
(a)
Give the trace \(f(x,2)\) at the fixed value \(y=2\) as a function of \(x\text{.}\) On Figure 11.3.7, draw the graph of the trace with \(y=2\) around the point where \(x=1\text{,}\) indicating the scale and labels on the axes. On the graph of the trace with \(y=2\text{,}\) sketch the tangent line at the point \(x=1\text{.}\)
Figure 11.3.7. A plot of \(f(x,2)\)
(b)
Find the partial derivative \(f_x(1,2)\) and relate its value to the sketch you just made.
(c)
Give the trace \(f(1,y)\) as a function of \(y\) at the fixed value \(x=1\text{.}\) On Figure 11.3.8, draw the graph of the trace with \(x=1\) indicating the scale and labels on the axes. On your graph of the trace corresponding to \(x=1\text{,}\) sketch the tangent line at the point \(y=2\text{.}\)
Figure 11.3.8. A plot of \(f(x,2)\)
(d)
Find the partial derivative \(f_y(1,2)\) and relate its value to the sketch you just made.
As these examples show, each partial derivative at a point arises as the derivative of a one-variable function defined by fixing all other coordinates. In addition, we may consider each partial derivative as defining a new function with input given by the point \((x,y)\text{,}\) just as the derivative \(f'(x)\) defines a new function of \(x\) in single-variable calculus. Due to the connection between one-variable derivatives and partial derivatives, we will often use Leibniz-style notation to denote partial derivatives by writing
\begin{equation*} \frac{\partial f}{\partial x}(a, b) = f_x(a,b), \ \mbox{and} \ \frac{\partial f}{\partial y}(a, b) = f_y(a,b). \end{equation*}
To calculate the partial derivative \(f_x\text{,}\) we hold \(y\) fixed and thus we treat \(y\) as a constant. In Leibniz notation, observe that
\begin{equation*} \frac{\partial }{\partial x} (x) = 1 \ \mbox{and} \ \frac{\partial }{\partial x}(y) = 0. \end{equation*}
To see the contrast between how we calculate single variable derivatives and partial derivatives, and the difference between the notations \(\frac{d}{dx}\bigl[ \ \bigr]\) and \(\frac{\partial}{\partial x}\bigl[ \ \bigr]\text{,}\) observe that
\begin{gather*} \frac{d}{dx}\Bigl[ 3x^2 - 2x + 3 \Bigr] = 3\frac{d}{dx} \Bigl[ x^2 \Bigr] - 2\frac{d}{dx} \Bigl[ x \Bigr] + \frac{d}{dx} \Bigl[ 3 \Bigr] = 3\cdot 2x - 2 \\ \frac{\partial}{\partial x}\Bigl[ x^2y - xy + 2y \Bigr] = y\frac{\partial}{\partial x} \Bigl[ x^2 \Bigr] - y\frac{\partial}{\partial x} \Bigl[ x \Bigr] + \frac{\partial}{\partial x} \Bigl[ 2y \Bigr] = y\cdot 2x - y \end{gather*}
We will highlight a few terms of the above calculations for emphasis. Note that when we are interested in taking the partial derivative with respect to \(x\) of a term like \(x^2y\text{,}\) we can factor out the \(y\) like it was a coefficient because \(y\) is constant with respect to changes in \(x\text{.}\) Algebraically, this looks like
\begin{equation*} \frac{\partial}{\partial x} \Bigl[ x^2y \Bigr] = y\frac{\partial}{\partial x} \Bigl[ x^2 \Bigr] \end{equation*}
Similarly, when we are taking the partial derivative with respect to \(x\) of a term like \(2y\text{,}\) the entire term is constant (with respect to \(x\)), so
\begin{equation*} \frac{\partial}{\partial x} \Bigl[ 2y \Bigr] = 0 \end{equation*}
Thus, computing partial derivatives is straightforward: we use the standard algebraic rules from single variable calculus, but do so while holding all but one of the variables constant.

Activity 11.3.3.
(a)
If \(f(x,y) = 3x^3 - 2x^2y^5\text{,}\) find the partial derivatives \(f_x\) and \(f_y\text{.}\)
(b)
If \(f(x,y) = \displaystyle\frac{xy^2}{x+1}\text{,}\) find the partial derivatives \(f_x\) and \(f_y\text{.}\)
(c)
If \(g(r,s) = rs\cos(r)\text{,}\) find the partial derivatives \(g_r\) and \(g_s\text{.}\)
(d)
Assuming \(f(w,x,y) = (6w+1)\cos(3x^2+4xy^3+y)\text{,}\) find the partial derivatives \(f_w\text{,}\) \(f_x\text{,}\) and \(f_y\text{.}\)
(e)
Find all possible first-order partial derivatives of \(q(x,t,z) = \displaystyle \frac{x2^tz^3}{1+x^2}.\)

Subsection 11.3.3 Interpretations of First-Order Partial Derivatives

Recall that the derivative of a single variable function has a geometric interpretation as the slope of the line tangent to the graph at a given point. Similarly, we have seen that the partial derivatives measure the slope of a line tangent to a trace of a function of two variables as shown in Figure 11.3.9.
Figure 11.3.9. A surface given by \(z=f(x,y)\) with traces \(y=b\) (in red) and \(x=a\) (in blue)
We will finish this section by looking at the meaning of partial derivatives as they relate to average rates of change for a function. Recall that the difference quotient \(\frac{f(a+h)-f(a)}{h}\) for a function \(f\) of a single variable \(x\) at a point where \(x=a\) tells us the average rate of change of \(f\) over the interval \([a,a+h]\text{,}\) while the derivative \(f'(a)\) tells us the instantaneous rate of change of \(f\) at \(x=a\text{.}\) We can use a similar relationship between limits of difference quotients and partial derivatives to provide context and meaning to partial derivatives as a measurement.

Activity 11.3.4.
The speed of sound \(C\) traveling through ocean water is a function of temperature, salinity and depth. It may be modeled by the function
\begin{align*} C \amp= 1449.2+4.6T-0.055T^2+0.00029T^3 \\ \amp \quad +(1.34-0.01T)(S-35)+0.016D \end{align*}
Here \(C\) is the speed of sound in meters/second, \(T\) is the temperature in degrees Celsius, \(S\) is the salinity in grams/liter of water, and \(D\) is the depth below the ocean surface in meters.
(a)
State the units for each of the partial derivatives, \(C_T\text{,}\) \(C_S\) and \(C_D\text{,}\) and explain the physical meaning of each partial derivative in terms of the rate of change for \(C\text{.}\)
(b)
Use algebraic rules to find the partial derivatives \(C_T\text{,}\) \(C_S\) and \(C_D\text{.}\)
(c)
Evaluate each of the three partial derivatives at the point where \(T=10\text{,}\) \(S=35\) and \(D=100\text{.}\) Write a few sentences to explain what the sign of each partial derivatives tell us about the behavior of the function \(C\) at the point \((10,35, 100)\text{.}\)

Subsection 11.3.4 Using Tables and Contour Plots to Estimate Partial Derivatives

Remember that functions of two variables can be represented numerical or graphically as either a table of data or a contour plot. In single variable calculus, if we only know the value of the function at a few points, then we can use the difference quotient to approximate derivatives. The same idea applies to partial derivatives; we need to make sure we are holding other input variables constant when we compute the difference quotient. The next two activities are focused on the meaning of partial derivatives using a table as a numerical representation of a multivariable function (Activity 11.3.4) and using a contour plot as a graphical representation of a multivariable function (Activity 11.3.5).

Activity 11.3.5.
The wind chill, as frequently reported, is a measure of how cold it feels outside when the wind is blowing. In Table 11.3.10, the wind chill \(w\text{,}\) measured in degrees Fahrenheit, is a function of the wind speed \(v\text{,}\) measured in miles per hour, and the ambient air temperature \(T\text{,}\) also measured in degrees Fahrenheit. We thus view \(w\) as being of the form \(w = w(v, T)\text{.}\)
Table 11.3.10. Wind chill as a function of wind speed and temperature
\(v \backslash T\) \(-20\) \(-15\) \(-10\) \(-5\) \(0\) \(5\) \(10\)
\(10\) \(-41\) \(-35\) \(-28\) \(-22\) \(-16\) \(-10\) \(-4\)
\(15\) \(-45\) \(-39\) \(-32\) \(-26\) \(-19\) \(-13\) \(-7\)
\(20\) \(-48\) \(-42\) \(-35\) \(-29\) \(-22\) \(-15\) \(-9\)
\(25\) \(-51\) \(-44\) \(-37\) \(-31\) \(-24\) \(-17\) \(-11\)
\(30\) \(-53\) \(-46\) \(-39\) \(-33\) \(-26\) \(-19\) \(-12\)
\(35\) \(-55\) \(-48\) \(-41\) \(-34\) \(-27\) \(-21\) \(-14\)
(a)
Do you think \(w_v\text{,}\) the partial derivative with respect to wind speed, is positive, negative, or zero at \((v,T)=(20,-10)\text{?}\) Write a couple of sentences to justify why you think the partial derivative is +/-/0 and remember to pay attention to which variable will be changing and which variable will be constant.
(b)
Do you think \(w_T\text{,}\) the partial derivative with respect to wind speed, is positive, negative, or zero at \((v,T)=(20,-10)\text{?}\) Write a couple of sentences to justify why you think the partial derivative is +/-/0 and remember to pay attention to which variable will be changing and which variable will be constant.
(c)
Estimate the partial derivative \(w_v(20,-10)\text{.}\) What are the units on this quantity and what does it mean? (Recall that we can estimate a partial derivative of a single variable function \(f\) using the symmetric difference quotient \(\frac{f(x+h)-f(x-h)}{2h}\) for small values of \(h\text{.}\) A partial derivative is a derivative of an appropriate trace.)
(d)
Estimate the partial derivative \(w_T(20,-10)\text{.}\) What are the units on this quantity and what does it mean?
(e)
Use your results to estimate the wind chill \(w(18, -10)\text{.}\) (Recall from single variable calculus that for a function \(f\) of \(x\text{,}\) \(f(x+h) \approx f(x) + hf'(x)\text{.}\))
(f)
Use your results to estimate the wind chill \(w(20, -12)\text{.}\)
(g)
Consider how you might combine your previous results to estimate the wind chill \(w(18, -12)\text{.}\) Explain your process.

Activity 11.3.6.
Shown below in Figure 11.3.11 is a contour plot of a function \(f\text{.}\) The values of the function on a few of the contours are indicated to the left of the figure.
Figure 11.3.11. A plot of \(f(t)=M(0.03,t)\)
(a)
Estimate the partial derivative \(f_x(-2,-1)\text{.}\) (Hint: How can you find values of \(f\) that are of the form \(f(-2+h)\) and \(f(-2-h)\) so that you can use a symmetric difference quotient?)
(b)
Estimate the partial derivative \(f_y(-2,-1)\text{.}\)
(c)
Estimate the partial derivatives \(f_x(-1,2)\) and \(f_y(-1,2)\text{.}\)
(d)
Locate, if possible, one point \((x,y)\) where \(f_x(x,y)= 0\text{.}\)
(e)
Locate, if possible, one point \((x,y)\) where \(f_x(x,y)\lt 0\text{.}\)
(f)
Locate, if possible, one point \((x,y)\) where \(f_y(x,y)>0\text{.}\)
Before we end this section, we will talk about a few ideas that were not necessary to understand and motivate the first partial derivatives, but will be important when we look at measuring or approximating the rate of change in other directions. In particular, we did not need to use our classic calculus approach to define the partial derivative because we restricted our two variable function along a trace to be a single variable function. This was extremely convenient but we will need new tools to make similar arguments in any other direction (that is not along a trace like \(x=a\) or \(y=b\)).
The approximations, algebraic rules, and interpretations for partial derivatives are the same as in single variable calculus. Additionally, the form of the difference quotients and limit definition for partial derivatives maintains the same form as in single variable calculus and for vector-valued functions of one variable:
\begin{equation*} \text{rate of change}= \lim \frac{\text{change in output of function}}{\text{change in input of function}} \end{equation*}
Again, conveniently all of the change we look at with a partial derivative is in a single input variable (holding all other input variables constant).

Subsection 11.3.5 Summary

  • If \(f=f(x,y)\) is a function of two variables, there are two first order partial derivatives of \(f\text{:}\) the partial derivative of \(f\) with respect to \(x\text{,}\)
    \begin{equation*} \frac{\partial f}{\partial x}(x,y) = f_x(x,y) = \lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}{h} \end{equation*}
    and the partial derivative of \(f\) with respect to \(y\)
    \begin{equation*} \frac{\partial f}{\partial y}(x,y) = f_y(x,y) = \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h}, \end{equation*}
    where each partial derivative exists only at those points \((x,y)\) for which the limit exists.
  • The partial derivative \(f_x(a,b)\) tells us the instantaneous rate of change of \(f\) with respect to \(x\) at \((x,y) = (a,b)\) when \(y\) is fixed at \(b\text{.}\) Geometrically, the partial derivative \(f_x(a,b)\) tells us the slope of the line tangent to the \(y=b\) trace of the function \(f\) at the point \((a,b,f(a,b))\text{.}\)
  • The partial derivative \(f_y(a,b)\) tells us the instantaneous rate of change of \(f\) with respect to \(y\) at \((x,y) = (a,b)\) when \(x\) is fixed at \(a\text{.}\) Geometrically, the partial derivative \(f_y(a,b)\) tells us the slope of the line tangent to the \(x=a\) trace of the function \(f\) at the point \((a,b,f(a,b))\text{.}\)

Exercises 11.3.6 Exercises

1.

Find the first partial derivatives of
\(\displaystyle f(x,y) = \frac{x - 4y}{x + 4y}\) at the point \((x,y) = (3, 3)\text{.}\)
\(\displaystyle \frac{\partial f}{\partial x}(3, 3) =\)
\(\displaystyle \frac{\partial f}{\partial y}(3, 3) =\)

2.

Find the first partial derivatives of \(f(x, y) = \sin(x - y)\) at the point (-4, -4).
A. \(f_x(-4, -4) =\)
B. \(f_y(-4, -4) =\)

3.

Find the partial derivatives of the function
\begin{equation*} w = \sqrt{4 r^2 + 2 s^2 + 5 t^2} \end{equation*}
\(\frac{\partial w}{\partial r} =\)
\(\frac{\partial w}{\partial s} =\)
\(\frac{\partial w}{\partial t} =\)

4.

Suppose that \(f(x,y)\) is a smooth function and that its partial derivatives have the values, \(f_x(0, 6) = 4\) and \(f_y(0, 6) = 4\text{.}\) Given that \(f(0, 6) = -1\text{,}\) use this information to estimate the value of \(f(1, 7)\text{.}\) Note this is analogous to finding the tangent line approximation to a function of one variable. In fancy terms, it is the first Taylor approximation.
Estimate of (integer value) \(f(0, 7)\)
Estimate of (integer value) \(f(1, 6)\)
Estimate of (integer value) \(f(1, 7)\)

5.

The gas law for a fixed mass \(m\) of an ideal gas at absolute temperature \(T\text{,}\) pressure \(P\text{,}\) and volume \(V\) is \(PV = mRT\text{,}\) where \(R\) is the gas constant.
Find the partial derivatives
\(\displaystyle \frac{\partial P}{\partial V} =\)
\(\displaystyle \frac{\partial V}{\partial T} =\)
\(\displaystyle \frac{\partial T}{\partial P} =\)
\(\displaystyle \frac{\partial P}{\partial V} \frac{\partial V}{\partial T}\frac{\partial T}{\partial P} =\) (an integer)

6.

Find the partial derivatives of the function
\begin{equation*} f(x,y) = \int_y^x \cos(4 t^2 - 7 t + 4)\, dt \end{equation*}
\(f_x(x,y) =\)
\(f_y(x,y) =\)

7.

Let \(f(x,y) = e^{-2x}\sin\!\left(5y\right)\text{.}\)
(a) Using difference quotients with \(\Delta x = 0.1\) and \(\Delta y = 0.1\text{,}\) we estimate
\(f_x (-3, -3) \approx\)
\(f_y (-3, -3) \approx\)
(b) Using difference quotients with \(\Delta x = 0.01\) and \(\Delta y = 0.01\text{,}\) we find better estimates:
\(f_x (-3, -3) \approx\)
\(f_y (-3, -3) \approx\)

8.

Determine the sign of \(f_x\) and \(f_y\) at each indicated point using the contour diagram of \(f\) shown below. (The point \(P\) is that in the first quadrant, at a positive \(x\) and \(y\) value; \(Q\) through \(T\) are located clockwise from \(P\text{,}\) so that \(Q\) is at a positive \(x\) value and negative \(y\text{,}\) etc.)
(a) At point \(P\text{,}\)
\(f_x\) is
and
\(f_y\) is
.
(b) At point \(Q\text{,}\)
\(f_x\) is
and
\(f_y\) is
.
(c) At point \(S\text{,}\)
\(f_x\) is
and
\(f_y\) is
.

9.

An experiment to measure the toxicity of formaldehyde yielded the data in the table below. The values show the percent, \(P=f(t,c)\text{,}\) of rats surviving an exposure to formaldehyde at a concentration of \(c\) (in parts per million, ppm) after \(t\) months.
\(t=14\) \(t=16\) \(t=18\) \(t=20\) \(t=22\) \(t=24\)
\(c=0\) 100 100 100 99 97 95
\(c=2\) 100 99 98 97 95 92
\(c=6\) 96 95 93 90 86 80
\(c=15\) 96 93 82 70 58 36
(a) Estimate \(f_t(16,2)\text{:}\)
\(f_t(16, 2)\approx\)
(b) Estimate \(f_c(16,2)\text{:}\)
\(f_c(16, 2)\approx\)
(Be sure that you can give the practical meaning of these two values in terms of formaldehyde toxicity.)

10.

An airport can be cleared of fog by heating the air. The amount of heat required depends on the air temperature and the wetness of the fog. The figure below shows the heat \(H(T,w)\) required (in calories per cubic meter of fog) as a function of the temperature \(T\) (in degrees Celsius) and the water content \(w\) (in grams per cubic meter of fog). Note that this figure is not a contour diagram, but shows cross-sections of \(H\) with \(w\) fixed at \(0.1\text{,}\) \(0.2\text{,}\) \(0.3\text{,}\) and \(0.4\text{.}\)
(a) Estimate \(H_T(10, 0.2)\text{:}\)
\(H_T(10,0.2) \approx\)
(Be sure you can interpret this partial derivative in practical terms.)
(b) Make a table of values for \(H(T,w)\) from the figure, and use it to estimate \(H_T(T,w)\) for each of the following:
\(T = 20, w = 0.2\) : \(H_T(T,w) \approx\)
\(T = 30, w = 0.2\) : \(H_T(T,w) \approx\)
\(T = 20, w = 0.3\) : \(H_T(T,w) \approx\)
\(T = 30, w = 0.3\) : \(H_T(T,w) \approx\)
(c) Repeat (b) to find \(H_w(T,w)\) for each of the following:
\(T = 20, w = 0.2\) : \(H_w(T,w) \approx\)
\(T = 30, w = 0.2\) : \(H_w(T,w) \approx\)
\(T = 20, w = 0.3\) : \(H_w(T,w) \approx\)
\(T = 30, w = 0.3\) : \(H_w(T,w) \approx\)
(Be sure you can interpret this partial derivative in practical terms.)

11.

The Heat Index, \(I\text{,}\) (measured in apparent degrees F) is a function of the actual temperature \(T\) outside (in degrees F) and the relative humidity \(H\) (measured as a percentage). A portion of the table which gives values for this function, \(I=I(T,H)\text{,}\) is reproduced in Table 11.3.12.
Table 11.3.12. A portion of the heat index data
T \(\downarrow \backslash\) H \(\rightarrow\) 70 75 80 85
90 106 109 112 115
92 112 115 119 123
94 118 122 127 132
96 125 130 135 141
  1. State the limit definition of the value \(I_T(94,75)\text{.}\) Then, estimate \(I_T(94,75)\text{,}\) and write one complete sentence that carefully explains the meaning of this value, including its units.
  2. State the limit definition of the value \(I_H(94,75)\text{.}\) Then, estimate \(I_H(94,75)\text{,}\) and write one complete sentence that carefully explains the meaning of this value, including its units.
  3. Suppose you are given that \(I_T(92,80) = 3.75\) and \(I_H(92,80) = 0.8\text{.}\) Estimate the values of \(I(91,80)\) and \(I(92,78)\text{.}\) Explain how the partial derivatives are relevant to your thinking.
  4. 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.

12.

Let \(f(x,y) = \frac{1}{2}xy^2\) represent the kinetic energy in Joules of an object of mass \(x\) in kilograms with velocity \(y\) in meters per second. Let \((a,b)\) be the point \((4,5)\) in the domain of \(f\text{.}\)
  1. Calculate \(f_x(a,b)\text{.}\)
  2. Explain as best you can in the context of kinetic energy what the partial derivative
    \begin{equation*} f_x(a,b) = \lim_{h \to 0} \frac{f(a+h,b) - f(a,b)}{h} \end{equation*}
    tells us about kinetic energy.
  3. Calculate \(f_y(a,b)\text{.}\)
  4. Explain as best you can in the context of kinetic energy what the partial derivative
    \begin{equation*} f_y(a,b) = \lim_{h \to 0} \frac{f(a,b+h) - f(a,b)}{h} \end{equation*}
    tells us about kinetic energy.
  5. Often we are given certain graphical information about a function instead of a rule. We can use that information to approximate partial derivatives. For example, suppose that we are given a contour plot of the kinetic energy function (as in Figure 11.3.13) instead of a formula. Use this contour plot to approximate \(f_x(4,5)\) and \(f_y(4,5)\) as best you can. Compare to your calculations from earlier parts of this exercise.
    Figure 11.3.13. The graph of \(f(x,y) = \frac{1}{2}xy^2\text{.}\)

13.

The temperature on an unevenly heated metal plate positioned in the first quadrant of the \(xy\)-plane is given by
\begin{equation*} C(x,y) = \frac{25xy+25}{(x-1)^2 + (y-1)^2 + 1}. \end{equation*}
Assume that temperature is measured in degrees Celsius and that \(x\) and \(y\) are each measured in inches. (Note: At no point in the following questions should you expand the denominator of \(C(x,y)\text{.}\))
  1. Determine \(\frac{\partial C}{\partial x}\restrict{(x,y)}\) and \(\frac{\partial C}{\partial y}\restrict{(x,y)}\text{.}\)
  2. If an ant is on the metal plate, standing at the point \((2,3)\text{,}\) and starts walking in the direction parallel to the positive \(y\) axis, at what rate will the temperature the ant is experiencing change? Explain, and include appropriate units.
  3. If an ant is walking along the line \(y = 3\) in the positive \(x\) direction, at what instantaneous rate will the temperature the ant is experiencing change when the ant passes the point \((1,3)\text{?}\)
  4. Now suppose the ant is stationed at the point \((6,3)\) and walks in a straight line towards the point \((2,0)\text{.}\) Determine the average rate of change in temperature (per unit distance traveled) the ant encounters in moving between these two points. Explain your reasoning carefully. What are the units on your answer?

14.

Consider the function \(f\) defined by \(f(x,y) = 8 - x^2 - 3y^2\text{.}\)
  1. Determine \(f_x(x,y)\) and \(f_y(x,y)\text{.}\)
  2. Find parametric equations in \(\R^3\) for the tangent line to the trace \(f(x,1)\) at \(x=2\text{.}\)
  3. Find parametric equations in \(\R^3\) for the tangent line to the trace \(f(2,y)\) at \(y=1\text{.}\)
  4. State respective direction vectors for the two lines determined in (b) and (c).
  5. Determine the equation of the plane that passes through the point \((2,1,f(2,1))\) whose normal vector is orthogonal to the direction vectors of the two lines found in (b) and (c).
  6. Use a graphing utility to plot both the surface \(z = 8 - x^2 - 3y^2\) and the plane from (e) near the point \((2,1)\text{.}\) What is the relationship between the surface and the plane?

15.

Recall from single variable calculus that, given the derivative of a single variable function and an initial condition, we can integrate to find the original function. We can sometimes use the same process for functions of more than one variable. For example, suppose that a function \(f\) satisfies \(f_x(x,y) = \cos(y)e^x+2x+y^2\text{,}\) \(f_y(x,y) = -\sin(y)e^x+2xy+3\text{,}\) and \(f(0,0) = 5\text{.}\)
  1. Find all possible functions \(f\) of \(x\) and \(y\) such that \(f_x(x,y) = \cos(y)e^x+2x+y^2\text{.}\) Your function will have both \(x\) and \(y\) as independent variables and may also contain summands that are functions of \(y\) alone.
  2. Use the fact that \(f_y(x,y) = -\sin(y)e^x+2xy+3\) to determine any unknown non-constant summands in your result from part (a).
  3. Complete the problem by determining the specific function \(f\) that satisfies the given conditions.