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Active Calculus - Multivariable

Section 11.4 Second-Order Partial Derivatives

Subsection 11.4.1 Introduction

Recall that for a single-variable function \(f\text{,}\) the second derivative of \(f\) is defined to be the derivative of the first derivative. That is, \(f''(x) = \frac{d}{dx}\Bigl[ f'(x) \Bigr] = \frac{d^2f}{dx^2}\text{,}\) which can be stated in terms of the limit definition of the derivative by writing
\begin{equation*} f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h}. \end{equation*}
In the following preview activity, we will explore how to calculate the four different second-order partial derivatives of a function of two variables as well as what these various derivatives tell us about the function’s behavior.

Preview Activity 11.4.1.
Once again, let’s consider the function \(f\) defined by \(f(x,y) = \frac{x^2\sin(2y)}{32}\) that measures a projectile’s range as a function of its initial speed \(x\) and launch angle \(y\text{.}\) The graph of this function, including traces with \(x=150\) and \(y=0.6\text{,}\) is shown in Figure 11.4.1.
Figure 11.4.1. The surface given by \(z=f(x,y) = \frac{x^2 \sin(2y)}{g}\) with traces \(y=0.6\) (in red) and \(x=150\) (in blue)
(a)
Compute the partial derivatives \(f_x\) and \(f_y\) as functions of \(x\) and \(y\text{.}\)
(b)
Notice that \(f_x\) itself is a new function of \(x\) and \(y\text{,}\) so we may now compute the partial derivatives of \(f_x\text{.}\) Find the partial derivative \(f_{xx} = (f_x)_x= \frac{\partial}{\partial x} \Bigl[ f_x \Bigr]\) and show that \(f_{xx}(150,0.6) \approx 0.058\text{.}\)
(c)
Figure 11.4.2 shows the trace of \(f\) with \(y=0.6\) with three tangent lines included. Write a few sentences to explain how \(f_{xx}(150,0.6) \approx 0.058\) is reflected in this figure.
Figure 11.4.2. A plot of \(f(x,0.6)\)
(d)
Find the partial derivative \(f_{yy}=(f_y)_y=\frac{\partial}{\partial y} \Bigl[ f_y \Bigr]\) and compute the value of \(f_{yy}(150, 0.6)\text{.}\)
(e)
Figure 11.4.3 shows the trace \(f(150, y)\) and includes three tangent lines. Write a couple of sentences to explain how the value of \(f_{yy}(150,0.6)\) is reflected in this figure.
Figure 11.4.3. A plot of \(f(x,0.6)\)
(f)
Because \(f_x\) and \(f_y\) are each functions of both \(x\) and \(y\text{,}\) they each have two partial derivatives. Not only can we compute \(f_{xx} = (f_x)_x\text{,}\) but also \(f_{xy} = (f_x)_y =\frac{\partial}{\partial y} \Bigl[ f_x \Bigr]\text{;}\) likewise, in addition to \(f_{yy} = (f_y)_y\text{,}\) but also \(f_{yx} = (f_y)_x =\frac{\partial}{\partial x} \Bigl[ f_y \Bigr]\text{.}\) For the range function \(f(x,y) = \frac{x^2\sin(2y)}{32}\text{,}\) use your earlier computations of \(f_x\) and \(f_y\) to now determine \(f_{xy}\) and \(f_{yx}\) (as functions of \(x\) and \(y\)). Write one sentence to explain how you calculated these “mixed” partial derivatives.
As we saw in the preview activity, we can relate properties of second derivatives from single variable calculus to the behavior of our trace functions. In this section, we will formalize our definitions of the second partial derivatives for a function of two variables and interpret how these give information about the functions behavior. Later in this section, we will briefly generalize to functions of more than two variables and higher order derivatives.

Subsection 11.4.2 Second-Order Partial Derivatives

A function \(f\) of two independent variables \(x\) and \(y\) has two first order partial derivatives, \(f_x=\frac{\partial f}{\partial x}\) and \(f_y=\frac{\partial f}{\partial y}\text{.}\) As we saw in Preview Activity 11.4.1, each of these first-order partial derivatives has two partial derivatives, giving a total of four second-order partial derivatives:
  • \(\displaystyle f_{xx} = (f_x)_x = \frac{\partial}{\partial x} \Bigl[ \frac{\partial f}{\partial x} \Bigr] = \frac{\partial^2 f}{\partial x^2}\)
  • \(\displaystyle f_{yy} = (f_y)_y=\frac{\partial}{\partial y} \Bigl[ \frac{\partial f}{\partial y} \Bigr] = \frac{\partial^2 f}{\partial y^2}\)
  • \(\displaystyle f_{xy} = (f_x)_y=\frac{\partial}{\partial y} \Bigl[ \frac{\partial f}{\partial x} \Bigr] = \frac{\partial^2 f}{\partial y \partial x}\)
  • \(\displaystyle f_{yx}=(f_y)_x=\frac{\partial}{\partial x} \Bigl[ \frac{\partial f}{\partial y} \Bigr] = \frac{\partial^2 f}{\partial x \partial y}\)
The first two, \(f_{xx}\) and \(f_{yy}\text{,}\) are called unmixed second-order partial derivatives while the last two, \(f_{xy}\) and \(f_{yx}\text{,}\) are called the mixed second-order partial derivatives.
One aspect of this notation can be a little confusing. The notation
\begin{equation*} \frac{\partial^2 f}{\partial y\partial x} = \frac{\partial}{\partial y} \Bigl[ \frac{\partial f}{\partial x} \Bigr] \end{equation*}
means that we first differentiate with respect to \(x\) and then with respect to \(y\text{;}\) this can be expressed in the alternate notation \(f_{xy} = (f_x)_y\text{.}\) However, to find the second partial derivative
\begin{equation*} f_{yx} = (f_y)_x \end{equation*}
we first differentiate with respect to \(y\) and then \(x\text{.}\) This means that
\begin{equation*} \frac{\partial^2 f}{\partial y\partial x} = f_{xy}, \ \mbox{and} \ \frac{\partial^2 f}{\partial x\partial y} = f_{yx}. \end{equation*}
Be sure to note carefully the difference between Leibniz notation and subscript notation and the order in which \(x\) and \(y\) appear in each. In addition, remember that anytime we compute a partial derivative, we hold constant the variable(s) other than the one we are differentiating with respect to.
In both notations, we use the same order of operations idea of working inside out. With the Leibniz notation, that means our partial derivative operators are written before the function we are taking partials of, whereas with the subscript notation writes order of derivatives after the function.

Activity 11.4.2.
(a)
Find all second order partial derivatives of the following functions. For each partial derivative you calculate, state explicitly which variable is being held constant.
  1. \(\displaystyle f(x,y) = x^2y^3\)
  2. \(\displaystyle y(q,p) = p\cos(q)\)
  3. \(\displaystyle g(s,t) = st^3 + s^4\)
(b)
If \(h(x,y,z,t) = 9x^9z-xyz^9 + 9t\text{,}\) how many second order partial derivatives does the function \(h\) have? Write a sentence to justify your reasoning on the number of second order partial derivatives of \(h\text{.}\) Finally, find \(h_{xz}\) and \(h_{zx}\) (you do not need to find the other second order partial derivatives).
In Preview Activity 11.4.1 and Activity 11.4.2, you may have noticed that the mixed second-order partial derivatives are equal. This observation holds generally and is known as Clairaut’s Theorem.

Clairaut’s Theorem.

Let \(f\) be a function of several variables for which the partial derivatives \(f_{xy}\) and \(f_{yx}\) are continuous near the point \((a,b)\text{.}\) Then
\begin{equation*} f_{xy}(a,b) = f_{yx}(a,b). \end{equation*}

Subsection 11.4.3 Interpreting the Second-Order Partial Derivatives

Recall from single variable calculus, that the first derivative describes whether a function is increasing or decreasing, but the second derivative measures how the function is increasing or decreasing.
Figure 11.4.4. Plots of three functions that are increasing, but doing so at an increasing rate, a constant rate, and a decreasing rate, respectively
In Figure 11.4.4, the left curve has the slopes of the tangent lines increasing (as you move along the graph from left to right) which means that the function is increasing at an increasing rate. In other words, the derivative of the slope of the tangent lines is positive, thus \(f''(x) \gt 0\text{.}\) We say that this curve is concave up. The middle curve in Figure 11.4.4 has tangent lines with the same slope, which means that the function is increasing at a constant rate. Thus the derivative of the slope of the tangent lines will be zero. The right curve in Figure 11.4.4 has the slope of the tangent lines decreasing (as you move along the graph from left to right) which means the function is increasing at a decreasing rate. In other words, the derivative of the slope of the tangent line is negative, so \(f''(x) \lt 0\text{.}\) We say that this curve is concave down.
A similar argument can be applied to the graphs of decreasing functions, as shown in Figure 11.4.5. In particular, we can say that the left plot will have negative slopes of tangent lines but those slopes are increasing (from left to right), so \(f''(x) \gt 0 \) and the curve is concave up. The middle plot in Figure 11.4.5 has a constant slope for tangent lines along the curve, thus the second derivative is zero. The right plot in Figure 11.4.5 has negative slopes of tangent lines (decreasing graph), and the slopes of those tangent lines are decreasing. Thus \(f''(x) \lt 0\) and we say that the graph is concave down.
Figure 11.4.5. Plots of three functions that are decreasing, but doing so at an increasing rate, a constant rate, and a decreasing rate, respectively
Remember that \(-10 \lt -5 \) and \(-5 \lt 2\text{,}\) so an increase in negative numbers means they are getting closer to zero and a decrease in negative numbers will mean that values are getting “more negative”.
Since the unmixed second-order partial derivative \(f_{xx}\) requires us to hold \(y\) constant and differentiate twice with respect to \(x\text{,}\) we can view \(f_{xx}\) as the second derivative of a trace of \(f\) where \(y\) is fixed. As such, \(f_{xx}\) will measure the concavity of this trace.

Example 11.4.6.

In this example, we will look at the second partial derivative of \(f(x,y) = \sin(x)e^{-y}\) with respect to \(x\) and how \(f_{xx}\) measures the concavity of the surface in the directions parallel to the \(x\)-axis.
Figure 11.4.7 shows a plot of the surface defined by \(z=f(x,y) = \sin(x)e^{-y}\) with the trace at \(y=-1.5\) highlighted (including a tangent line). You can use the slider at the top of Figure 11.4.7 to change the location of the tangent line shown on the trace given by \(y=-1.5\text{.}\) As with ALL plots in three dimensions, you should look at the axes or the coordinate values shown in the frame of the plot to make sure you understand which direction corresponds to an increase in each variable. This is especially important with interactive plots like the ones used throughout this text because you may have changed the perspective from the original settings in order to see the plot from a different view.
As you increase the \(x\)-coordinate where the tangent line is shown, you should notice that the slope of the tangent line is decreasing, and just as in single variable calculus, this is why the trace is concave down.
Figure 11.4.7. The surface given by \(z=f(x,y) = \sin(x)e^{-y}\) with trace at \(y=-1.5\) and tangent line drawn
Algebraically, we can calculate \(f_x(x,y)=\cos(x)e^{-y}\) and so \(f_{xx}(x,y)=-\sin(x)e^{-y} \text{.}\) If we consider \(x\) coordinates from \(0\) to \(\pi\text{,}\) then \(f_{xx}(x,y)=-\sin(x)e^{-y} \lt 0 \) since \(-\sin(x) \leq 0\) if \(x\in[0,\pi]\) and \(e^{-y} \gt 0\) for all values of \(y\text{,}\) including \(y = -1.5\text{.}\) The other second partials of \(f\) are explored in Activity 11.4.3.
In the following activity, we further explore what second-order partial derivatives tell us about the geometric behavior of a surface.

Activity 11.4.3.
We will continue to work with the function \(f\) defined by \(f(x,y) = \sin(x) e^{-y}\text{.}\)
(a)
In Figure 11.4.8, we see the trace of \(f(x,y) = \sin(x) e^{-y}\) with \(x\) held constant with \(x = 1.75\) plotted in blue. You can use the slider in Figure 11.4.8 to look at how the slope of the tangent line changes as you vary the \(y\)-coordinate along this trace. Write a couple of sentences that describe whether the slope of the tangent lines to this curve increase or decrease as \(y\) increases along the trace \(x=1.75\text{.}\) Be sure to pay attention to which direction corresponds to each coordinate increasing.
Figure 11.4.8. The surface given by \(z=f(x,y) = \sin(x)e^{-y}\) with trace at \(x=1.75\) and tangent line drawn
(b)
Compute \(f_{yy}(x,y)\) algebraically and explain how your observations in the previous task are related to the value of \(f_{yy}(1.75,y)\text{.}\) Be sure to address the notion of concavity in your response and be careful to note the directions in which \(y\) is increasing.
(c)
We want to explore what is described by the mixed partial derivative, \(f_{xy}\text{,}\) and we will focus on the point given by \((x,y)=(1.75,-1.5)\text{.}\) In this task, we will work through the definition for \(f_{xy} (1.75,-1.5)\) carefully.
The partial derivative \(f_x(1.75,-1.5)\) measures the slope of the line tangent to the trace given by \(y=-1.5\text{.}\) The tangent line we are examining is changing in the \(x\)-direction (parallel to the \(x\)-axis). When we look at \(f_{xy}\text{,}\) we are taking the partial derivative of \(f_x\) with respect to \(y\text{,}\) \(\frac{\partial}{\partial y} \Bigl[ f_x \Bigr]\text{.}\) So we are looking at how the slope of the tangent line in the \(x\)-direction will change when we vary the \(y\)-coordinate a little bit. We would approximate \(f_{xy}(1.75,-1.5)\) with the following difference quotient:
\begin{equation*} \frac{f_x(1.75,-1.5+h)-f_x(1.75,-1.5)}{h} \end{equation*}
You might think of sliding your pencil down the trace with \(x=1.5\) so that the pencil is in the \(x\)-direction and tangent to the surface. In Figure 11.4.9, the pencil would be the black tangent line drawn. You can use the slider at the top of Figure 11.4.9 to change the \(y\)-coordinate of the point where the black tangent line is drawn. You should look at what happens to the slope of the black tangent line as you increase the \(y\)-coordinate with the slider. The tangent line in the \(x\)-direction at the point \((1.75,-1.5)\) is drawn in gray for reference purposes.
Based on your exploration with Figure 11.4.9, write a few sentences about whether \(f_{xy}(1.75, -1.5)\) is positive or negative and justify your reasoning.
Figure 11.4.9. The surface given by \(z=f(x,y) = \sin(x)e^{-y}\) with trace at \(x=1.75\) and a tangent line in the \(x\)-direction drawn
(d)
Compute \(f_{xy}(x,y)\) algebraically and evaluate \(f_{xy}(1.75, -1.5)\text{.}\) Write a couple of sentences about how this value compares with your observations in the previous task.
(e)
We know that \(f_{xx}(1.75, -1.5)\) measures the concavity of the \(y = -1.5\) trace, and that \(f_{yy}(1.75, -1.5)\) measures the concavity of the \(x = 1.75\) trace. What do you think the quantity \(f_{xy}(1.75, -1.5)\) measures?
(f)
On Figure 11.4.10, the trace with \(y = -1.5\) is highlighted with the point \((1.75,-1.5,f(1.75,-1.5))\) drawn in black. Sketch three tangent lines whose slopes correspond to the value of \(f_{yx}(x,-1.5)\) for three different values of \(x\) (near \(x=1.75\)). Use your tangent lines to state whether \(f_{yx}(1.75, -1.5)\) is positive or negative. Justify your reasoning and describe what you think \(f_{yx}(1.75, -1.5)\) measures.
Figure 11.4.10. The surface given by \(z=f(x,y) = \sin(x)e^{-y}\) with trace at \(y=-1.5\)
As we have found in Activities 11.4.3 and Activity 11.4.4, we may think of \(f_{xy}\) as measuring the “twist” of the graph as we increase \(y\) along a particular trace where \(x\) is held constant. In the same way, \(f_{yx}\) measures how the graph twists as we increase \(x\text{.}\) If we remember that Clairaut’s theorem tells us that \(f_{xy} = f_{yx}\text{,}\) we see that the amount of twisting is the same in both directions. This twisting is perhaps more easily seen in Figure 11.4.11, which shows the graph of \(f(x,y) = -xy\text{,}\) for which \(f_{xy} = -1\text{.}\)
Figure 11.4.11. The surface given by \(z=f(x,y) = xy\)
Just as with the first-order partial derivatives, we can approximate second-order partial derivatives in the situation where we have information about the function only at certain inputs, like with tables or contour plots.

Activity 11.4.4.
As we saw in Activity 11.3.5, the wind chill \(w(v,T)\text{,}\) in degrees Fahrenheit, is a function of the wind speed, in miles per hour, and the air temperature, in degrees Fahrenheit. Some values of the wind chill are recorded in Table 11.4.12.
Table 11.4.12. 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)
Estimate the partial derivatives \(w_{T}(20,-15)\text{,}\) \(w_{T}(20,-10)\text{,}\) and \(w_T(20,-5)\text{.}\) Use these results to estimate the second-order partial \(w_{TT}(20, -10)\text{.}\)
(b)
In a similar way, estimate the second-order partial \(w_{vv}(20,-10)\text{.}\)
(c)
Estimate the partial derivatives \(w_T(20,-10)\text{,}\) \(w_T(25,-10)\text{,}\) and \(w_T(15,-10)\text{,}\) and use your results to estimate the partial \(w_{Tv}(20,-10)\text{.}\)
(d)
In a similar way, estimate the partial derivative \(w_{vT}(20,-10)\text{.}\)
(e)
Write a few sentences to explain what the values \(w_{TT}(20, -10)\text{,}\) \(w_{vv}(20,-10)\text{,}\) and \(w_{Tv}(20,-10)\) indicate regarding the behavior of \(w(v,T)\text{.}\)

Subsection 11.4.4 Summary

  • There are four second-order partial derivatives of a function \(f\) of two independent variables \(x\) and \(y\text{:}\)
    \begin{equation*} f_{xx} = (f_x)_x, f_{xy} = (f_x)_y, f_{yx} = (f_y)_x,\ \mbox{and} \ f_{yy} = (f_y)_y. \end{equation*}
  • The unmixed second-order partial derivatives, \(f_{xx}\) and \(f_{yy}\text{,}\) tell us about the concavity of the traces. The mixed second-order partial derivatives, \(f_{xy}\) and \(f_{yx}\text{,}\) tell us how the graph of \(f\) twists.

Exercises 11.4.5 Exercises

1.

Calculate all four second-order partial derivatives of \(\displaystyle f(x,y) = 3x^{2}y+5xy^{3}\text{.}\)
\(f_{xx} \, (x,y) =\)
\(f_{xy} \, (x,y) =\)
\(f_{yx} \, (x,y) =\)
\(f_{yy} \, (x,y) =\)

2.

Find all the first and second order partial derivatives of \(f(x, y) = -2\sin(2x+y) - 10\cos(x-y)\text{.}\)
A. \(\frac{\partial f}{\partial x} = f_x =\)
B. \(\frac{\partial f}{\partial y} = f_y =\)
C. \(\frac{{\partial^2}f}{\partial x^2} = f_x{}_x =\)
D. \(\frac{{\partial^2}f}{\partial y^2} = f_y{}_y =\)
E. \(\frac{{\partial^2}f}{\partial x \partial y} = f_y{}_x =\)
F. \(\frac{{\partial^2}f}{\partial y \partial x} = f_x{}_y =\)

3.

Find the partial derivatives of the function
\begin{equation*} f(x,y) = xye^{4 y} \end{equation*}
\(f_x(x,y) =\)
\(f_y(x,y) =\)
\(f_{xy}(x,y) =\)
\(f_{yx}(x,y) =\)

4.

Calculate all four second-order partial derivatives of \(\displaystyle f(x,y) = \sin\!\left(\frac{4x}{y}\right)\text{.}\)
\(f_{xx} \, (x,y) =\)
\(f_{xy} \, (x,y) =\)
\(f_{yx} \, (x,y) =\)
\(f_{yy} \, (x,y) =\)

5.

Given \(F(r,s,t)=r\!\left(4s^{4}-9t^{6}\right)\text{,}\) compute:
\(F_{rst}=\)

6.

Calculate all four second-order partial derivatives and check that \(f_{xy}=f_{yx}\text{.}\) Assume the variables are restricted to a domain on which the function is defined.
\begin{equation*} f(x,y) = e^{5xy} \end{equation*}
\(f_{xx} =\)
\(f_{yy} =\)
\(f_{xy} =\)
\(f_{yx} =\)

7.

Calculate all four second-order partial derivatives of \(\displaystyle f(x,y) = \left(5x+3y\right)e^{y}\text{.}\)
\(f_{xx} \, (x,y) =\)
\(f_{xy} \, (x,y) =\)
\(f_{yx} \, (x,y) =\)
\(f_{yy} \, (x,y) =\)

8.

Let \(f(x,y) = \left(-\left(5x+y\right)\right)^{9}\text{.}\) Then
\(\frac{\partial^2\!f}{\partial x\partial y}\) \(=\)
\(\frac{\partial^3\!f}{\partial x\partial y\partial x}\) \(=\)
\(\frac{\partial^3\!f}{\partial x^2\partial y}\) \(=\)

9.

If \(z_{xy} = 4 y\) and all of the second order partial derivatives of \(z\) are continuous, then
(a) \(z_{yx} =\)
(b) \(z_{xyx} =\)
(c) \(z_{xyy} =\)

10.

If \(z = f(x) + y g(x)\text{,}\) what can we say about \(z_{yy}\text{?}\)

11.

Shown in Figure 11.4.13 is a contour plot of a function \(f\) with the values of \(f\) labeled on the contours. The point \((2,1)\) is highlighted in red.
Figure 11.4.13. A contour plot of \(f(x,y)\text{.}\)
  1. Estimate the partial derivatives \(f_x(2,1)\) and \(f_y(2,1)\text{.}\)
  2. Determine whether the second-order partial derivative \(f_{xx}(2,1)\) is positive or negative, and explain your thinking.
  3. Determine whether the second-order partial derivative \(f_{yy}(2,1)\) is positive or negative, and explain your thinking.
  4. Determine whether the second-order partial derivative \(f_{xy}(2,1)\) is positive or negative, and explain your thinking.
  5. Determine whether the second-order partial derivative \(f_{yx}(2,1)\) is positive or negative, and explain your thinking.
  6. Consider a function \(g\) of the variables \(x\) and \(y\) for which \(g_x(2,2) > 0\) and \(g_{xx}(2,2) \lt 0\text{.}\) Sketch possible behavior of some contours around \((2,2)\) on the left axes in Figure 11.4.14.
    Figure 11.4.14. Plots for contours of \(g\) and \(h\text{.}\)
  7. Consider a function \(h\) of the variables \(x\) and \(y\) for which \(h_x(2,2) > 0\) and \(h_{xy}(2,2) \lt 0\text{.}\) Sketch possible behavior of some contour lines around \((2,2)\) on the right axes in Figure 11.4.14.

12.

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, \(I(T,H)\text{,}\) is reproduced in Table 11.4.15.
Table 11.4.15. Heat index.
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 \(I_{TT}(94,75)\text{.}\) Then, estimate \(I_{TT}(94,75)\text{,}\) and write one complete sentence that carefully explains the meaning of this value, including units.
  2. State the limit definition of the value \(I_{HH}(94,75)\text{.}\) Then, estimate \(I_{HH}(94,75)\text{,}\) and write one complete sentence that carefully explains the meaning of this value, including units.
  3. Finally, do likewise to estimate \(I_{HT}(94,75)\text{,}\) and write a sentence to explain the meaning of the value you found.

13.

The temperature on a heated metal plate positioned in the first quadrant of the \(xy\)-plane is given by
\begin{equation*} C(x,y) = 25e^{-(x-1)^2 - (y-1)^3}. \end{equation*}
Assume that temperature is measured in degrees Celsius and that \(x\) and \(y\) are each measured in inches.
  1. Determine \(C_{xx}(x,y)\) and \(C_{yy}(x,y)\text{.}\) Do not do any additional work to algebraically simplify your results.
  2. Calculate \(C_{xx}(1.1, 1.2)\text{.}\) Suppose that an ant is walking past the point \((1.1, 1.2)\) along the line \(y = 1.2\text{.}\) Write a sentence to explain the meaning of the value of \(C_{xx}(1.1, 1.2)\text{,}\) including units.
  3. Calculate \(C_{yy}(1.1, 1.2)\text{.}\) Suppose instead that an ant is walking past the point \((1.1, 1.2)\) along the line \(x = 1.1\text{.}\) Write a sentence to explain the meaning of the value of \(C_{yy}(1.1, 1.2)\text{,}\) including units.
  4. Determine \(C_{xy}(x,y)\) and hence compute \(C_{xy}(1.1, 1.2)\text{.}\) What is the meaning of this value? Explain, in terms of an ant walking on the heated metal plate.

14.

Let \(f(x,y) = 8 - x^2 - y^2\) and \(g(x,y) = 8 - x^2 + 4xy - y^2\text{.}\)
  1. Determine \(f_x\text{,}\) \(f_y\text{,}\) \(f_{xx}\text{,}\) \(f_{yy}\text{,}\) \(f_{xy}\text{,}\) and \(f_{yx}\text{.}\)
  2. Evaluate each of the partial derivatives in (a) at the point \((0,0)\text{.}\)
  3. What do the values in (b) suggest about the behavior of \(f\) near \((0,0)\text{?}\) Plot a graph of \(f\) and compare what you see visually to what the values suggest.
  4. Determine \(g_x\text{,}\) \(g_y\text{,}\) \(g_{xx}\text{,}\) \(g_{yy}\text{,}\) \(g_{xy}\text{,}\) and \(g_{yx}\text{.}\)
  5. Evaluate each of the partial derivatives in (d) at the point \((0,0)\text{.}\)
  6. What do the values in (e) suggest about the behavior of \(g\) near \((0,0)\text{?}\) Plot a graph of \(g\) and compare what you see visually to what the values suggest.
  7. What do the functions \(f\) and \(g\) have in common at \((0,0)\text{?}\) What is different? What do your observations tell you regarding the importance of a certain second-order partial derivative?

15.

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 \(\frac{ \partial^2 f}{\partial x^2}\) at the point \((a,b)\text{.}\) Then explain as best you can what this second order partial derivative tells us about kinetic energy.
  2. Calculate \(\frac{ \partial^2 f}{\partial y^2}\) at the point \((a,b)\text{.}\) Then explain as best you can what this second order partial derivative tells us about kinetic energy.
  3. Calculate \(\frac{ \partial^2 f}{\partial y \partial x}\) at the point \((a,b)\text{.}\) Then explain as best you can what this second order partial derivative tells us about kinetic energy.
  4. Calculate \(\frac{ \partial^2 f}{\partial x \partial y}\) at the point \((a,b)\text{.}\) Then explain as best you can what this second order partial derivative tells us about kinetic energy.