What do the second-order partial derivatives \(f_{xx}\text{,}\)\(f_{yy}\text{,}\)\(f_{xy}\text{,}\) and \(f_{yx}\) of a function \(f\) tell us about the function’s behavior?
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
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.
Once again, we 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 the interactive graph below.
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 \(\frac{\partial}{\partial x} \Bigl[ f_x \Bigr]\text{,}\) which we will denote by \(f_{xx} \text{.}\) Verify that \(f_{xx}(150,0.6) \approx 0.058\text{.}\)
The graph below 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.
Find the partial derivative \(\frac{\partial}{\partial y} \Bigl[ f_y \Bigr]\text{,}\) which we will denote by \(f_{yy}\) and compute the value of \(f_{yy}(150, 0.6)\text{.}\)
The graph below 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.
Because \(f_x\) and \(f_y\) are each functions of both\(x\) and \(y\text{,}\) they each have two partial derivatives. You have already computed two of them. Now compute
for the range function \(f(x,y) = \frac{x^2\sin(2y)}{32}\) by using your earlier computations of \(f_x\) and \(f_y\text{.}\) Write a 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.
A function \(f\) of two 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:
We call \(f_{xx}\) and \(f_{yy}\) unmixed second-order partial derivatives. We refer to \(f_{xy}\) and \(f_{yx}\) as the mixed second-order partial derivatives.
we mean that we first differentiate with respect to \(x\) and then differentiate 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
The Leibniz notation and subscript notation for a mixed partial differ in terms of the order in which \(x\) and \(y\) appear in each notation. However, in both notations, we use the idea of order of operations working from the inside out. With the Leibniz notation, that means our partial derivative operators are written before the function we are taking partial derivatives of, whereas with the subscript notation writes order of derivatives after the function.
Find all second order partial derivatives of the following functions. For each partial derivative you calculate, state explicitly which variable is being held constant.
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.
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
Subsection11.4.3Interpreting 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.
In Figure 11.4.2, 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, making \(f''(x) \gt 0\text{.}\) We say that this curve is concave up. The middle curve in Figure 11.4.2 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 11.4.2 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 11.4.3. 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 11.4.3 has a constant slope for tangent lines along the curve, thus the second derivative is zero. The right plot in Figure 11.4.3 has negative slopes of tangent lines, corresponding to a 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.
Remember that \(-10 \lt -5 \) and \(-5 \lt 2\text{,}\) so an increase in negative numbers means the numbers are getting closer to zero, while a decrease in negative numbers means that values are getting “more negative” or farther from zero.
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.
In this example, we will look at the second-order unmixed 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 11.4.5 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). Move the slider 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 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 decreases, and just as in single variable calculus, this is why the trace is concave down.
Algebraically, we can calculate \(f_x(x,y)=\cos(x)e^{-y}\) and \(f_{xx}(x,y)=-\sin(x)e^{-y} \text{.}\) 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 \(0\leq x\leq \pi\) 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 interactive graphic below we see the trace of \(f(x,y) = \sin(x) e^{-y}\) with \(x\) held constant with \(x = 1.75\) plotted in blue. Use the slider to investigate 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.
Compute \(f_{yy}(x,y)\) algebraically and explain how your observations in the previous part are related to the value of \(f_{yy}(1.75,y)\text{.}\) Your response should address the notion of concavity. Be careful to note the directions in which \(y\) is increasing.
We want to explore what the mixed partial derivative \(f_{xy}\) describes. We will do this by focusing on the point \((x,y)=(1.75,-1.5)\text{.}\) In this part, 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 \(y=-1.5\text{.}\) These tangent lines change in the \(x\)-direction (parallel to the \(x\)-axis). When we consider \(f_{xy}\text{,}\) we are taking the partial derivative of \(f_x\) with respect to \(y\text{,}\)\(\frac{\partial}{\partial y} \Bigl[ f_x \Bigr]\text{.}\) Thus, we are considering how the slope of the tangent line in the \(x\)-direction changes when we vary the \(y\)-coordinate a small amount. We can approximate \(f_{xy}(1.75,-1.5)\) with the difference quotient
One way of visualizing this is by thinking of sliding a pencil along the trace with \(x=1.5\) so that the pencil is in the \(x\)-direction and tangent to the surface. In the interactive image below, the pencil would be the black tangent line drawn. Use the slider at the top of the interactive to change the \(y\)-coordinate of the point where the tangent line is drawn. Examine what happens to the slope of the tangent line as you increase the \(y\)-coordinate of the point of tangency by adjusting the slider. The tangent line in the \(x\)-direction at the point \((1.75,-1.5)\) is drawn in gray for reference purposes.
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 part.
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?
In the interactive image below, 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 to this trace whose slopes correspond to the value of \(f_{yx}(x,-1.5)\) for three different values of \(x\) near \(x=1.75\text{.}\) 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.
As we have seen in Activity 11.4.3, 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 since \(f\) is continuous, \(f_{xy} = f_{yx}\text{,}\) we see that the amount of twisting is the same in both directions. This twisting is perhaps more easily seen in Figure 11.4.6, which shows the graph of \(f(x,y) = -xy\text{,}\) for which \(f_{xy} = -1\text{.}\)
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.
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 the table below.
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 \(w_{TT}(20, -10)\text{.}\)
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 \(w_{Tv}(20,-10)\text{.}\)
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 \(w(v,T)\text{.}\)
The unmixed second-order partial derivatives \(f_{xx}\) and \(f_{yy}\) tell us about the concavity of the traces. The mixed second-order partial derivatives \(f_{xy}\) and \(f_{yx}\) tell us how the graph of \(f\) twists.
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.
Shown in Figure 11.4.7 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.
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 11.4.8.
Figure11.4.8.Plots for contours of \(g\) and \(h\text{.}\)
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 11.4.8.
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 11.4.9.
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.
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.
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.
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.
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.
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.
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.
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?
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 \(f\text{.}\)
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.
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.
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.
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.
