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Active Calculus - Multivariable

Section 11.6 The Multivariable Chain Rule

Subsection 11.6.1 Introduction

In single-variable calculus, we encountered situations in which some quantity was related through a composition of related quantities. For instance, let \(P\) be the price of a product and \(P\) depends on \(x\text{,}\) the wages of people making the product. The wages of people making the product will also change over time, denoted \(t\text{,}\) because of seasonal availability and other factors. So a change in the value of \(t\) will produce a change in the value of \(x\text{,}\) which will produce a change in the value of \(P\text{.}\) This means that we can write \(P\) as a function of \(x\) and \(x\) as a function of \(t\text{,}\) thus we can express \(P\)’s dependence on \(t\) by a composition of functions:
Figure 11.6.1. Composition of \(P(x)\) and \(x(t)\)
Using the language of differentials that we saw in the previous section and the ideas of local linearity we can express these changes as follows. For a small change in \(x\text{,}\) the change in the price of the product will be approximately proportional to the change in \(x\)
\begin{equation*} dP = (m) dx \end{equation*}
with \(m\) being the value of \(\frac{dP}{dx}\) at \(x_0\text{,}\) the current value of \(x\text{.}\) We can write \(v\) as \(m=\frac{dP}{dx}\restrict{x=x_0}\) because \(m\) is current instantaneous rate of change in the product price per unit of labor cost.
A small change in the value of \(t\) will also cause a proportional change in the value of \(x\text{.}\)
\begin{equation*} dx = (v) dt \end{equation*}
with \(v\) being the value of \(\frac{dx}{dt}\) at the current value of \(t\text{,}\) which we will call \(t_0\text{.}\) So \(v=\frac{dx}{dt}\restrict{t=t_0}\text{.}\)
We can now combine these relationships to express how a change in \(t\) will produce a change in \(P\text{.}\)
\begin{equation*} dP=(m) dx = (m)(v)dt = \left( \frac{dP}{dx}\restrict{x=x_0} \right)\left( \frac{dx}{dt}\restrict{t=t_0} \right) dt \end{equation*}
In particular, this says that a change in the price of our product will the current instantaneous rate of change of the price per step in \(x\) times the current rate of change in the price of labor times the step in time.
The previous equation is called the Chain Rule. In single variable calculus, you likely saw the Chain Rule stated as: If \(C\) is the composition of functions \(f\) and \(g\) such \(C(x)=f(g(x))\text{,}\) \(f\) is differentiable at \(g(x)\text{,}\) and \(g\) is differentiable at \(x\text{,}\) then
\begin{equation*} C'(x)=f'(g(x)) g'(x) \end{equation*}
If we look at a function of two variables where \(z = f(x,y)\text{,}\) it might be that both \(x\) and \(y\) depend on another variable \(t\text{.}\) A change in \(t\) then produces changes in both \(x\) and \(y\text{,}\) which then causes \(z\) to change. In this section, we will see how to describe the change in \(z\) that is caused by a change in \(t\text{,}\) leading us to multivariable versions of the Chain Rule involving both regular and partial derivatives.

Preview Activity 11.6.1.
Your self-driving car company, Steer Clear, is doing well and almost ready to launch its first car. Some of your engineers have reported that the car has problems when the air intake encounters too much large particulate matter in the air (like sand, dust, large pollen, smog, etc.). In order to fix this issue, you have created a new type of filter that uses a sophisticated mesh and gravity to filter out large particulate matter, but this new type of filter gets clogged if exposed to too much large particulate matter too quickly. So we will need to measure the rate at which your car’s intake will be exposed to large particulate matter per unit time.
You consult a friend Alex who does atmospheric modeling of large particulate matter in your area. Alex created \(P(x,y)=10-\frac{1}{2} x^2-\frac{1}{5}y^2\text{,}\) a function that describes the amount of large particular matter in the air in terms of location relative to Alex’s lab. In particular, the \(x\) coordinate is the distance East/West of from Alex’s lab in kilometers and the \(y\)-coordinate is the distance North/South from Alex’s lab in kilometers.
Since you have completed the self-driving part of your car, you know that your car will move along a test course as a function of time given by
\begin{equation*} \vr(t) = \langle x(t), y(t) \rangle = \langle 2-t^2, t^3 + 1\rangle. \end{equation*}
where the \(x\) and \(y\) coordinates are the same as measured by Alex’s function and \(t\) is measured in minutes.
(a)
Substitute \(x(t)\) and \(y(t)\) (from your self-driving car path) into your expression for \(P(x,y)\) to get an expression for \(P(t)\text{,}\) the amount of large particulate matter encountered at each location on your test course as a function of time driven. Do not simplify your expression for \(P(t)\text{.}\)
(b)
Use your derivative rules from single variable calculus to find \(P'(t)=\frac{dP}{dt}\text{.}\) Note that this is not a partial derivative because \(P(t)\) only has one dependent variable. You should NOT simplify your expression for \(P'(t)\text{.}\)
(c)
Alex has told you that the amount of large particulate matter in the air changes in the \(y\)-direction (North/South) because of the proximity to industrial pollution from a factory complex. Also, the amount of large particulate matter in the air changes in the \(x\)-direction (East/West) because of tree pollen from a particular kind of tree in a nearby forest. We would like to understand how the rate of large particulate matter encountered on our drive is split into these two factor (industrial pollution and tree pollen).
Alex suggests that you measure \(\frac{dP}{dt}\) at a location \((x,y)=(a,b)\) by taking the derivative of the linearization for \(P\) at \((a,b)\text{.}\) The linearization of \(P\) at \((a,b)\) is given by
\begin{equation*} L(x,y)=P(a,b)+P_x(a,b)(x-a)+P_y(a,b)(y-b) \end{equation*}
Remember that we will need to treat \(x\) and \(y\) as functions of time.
Write a few sentences to explain why
\begin{equation} \frac{dP}{dt}\approx\frac{dL}{dt} = 0 + P_x(a,b) \frac{dx}{dt}+P_y(a,b) \frac{dy}{dt}\tag{11.6.1} \end{equation}
and how each of \(P_x(a,b) \frac{dx}{dt}\) and \(P_y(a,b) \frac{dy}{dt}\) measure the rate of change in pollution coming from tree pollen and industrial pollution.
(d)
Compute each of the following and substitute your expressions into (11.6.1), Alex’s approximation. Do NOT simplify your expression, just substitute in the values for each element below.
  • \(\displaystyle P_x(a,b) \)
  • \(\displaystyle P_x(a,b) \)
  • \(\displaystyle \frac{dx}{dt} \)
  • \(\displaystyle \frac{dy}{dt} \)
(e)
Compare your results for part 11.6.1.b and part 11.6.1.d and write a few sentences that identifies how each part of your answer for part 11.6.1.d corresponds to different parts of part 11.6.1.b. Remember your work for part 11.6.1.b is completely in terms of \(t\) so you will need to translate some terms from \(t\) into its meaning in \(x\text{,}\) \(y\text{,}\) or others.

Subsection 11.6.2 The Chain Rule

As Preview Activity 11.4.1 suggests, the following version of the Chain Rule holds in general.

The Chain Rule.

Let \(z = f(x,y)\text{,}\) where \(f\) is a differentiable function of the independent variables \(x\) and \(y\text{,}\) and let \(x\) and \(y\) each be differentiable functions of an independent variable \(t\text{.}\) Then
\begin{equation} \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}.\tag{11.6.2} \end{equation}
It is important to note the differences among the derivatives in (11.6.2). Since \(z\) is a function of the two variables \(x\) and \(y\text{,}\) the derivatives in the Chain Rule for \(z\) with respect to \(x\) and \(y\) are partial derivatives. However, since \(x = x(t)\) and \(y = y(t)\) are functions of the single variable \(t\text{,}\) their derivatives are the standard derivatives of functions of one variable. When we compose \(z\) with \(x(t)\) and \(y(t)\text{,}\) we then have \(z\) as a function of the single variable \(t\text{,}\) making the derivative of \(z\) with respect to \(t\) a standard derivative from single variable calculus as well.
To understand why this Chain Rule works in general, suppose that some quantity \(z\) depends on \(x\) and \(y\) so that we can express the change in \(z\) in terms of the differential as
\begin{equation} dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy.\tag{11.6.3} \end{equation}
Next, suppose that \(x\) and \(y\) each depend on another quantity \(t\text{,}\) so that
\begin{equation} dx = \frac{dx}{dt}~dt \ \mbox{and} \ dy = \frac{dy}{dt}~dt.\tag{11.6.4} \end{equation}
Combining Equations (11.6.3) and (11.6.4), we find that
\begin{equation*} dz = \frac{\partial z}{\partial x}\frac{dx}{dt}~dt + \frac{\partial z}{\partial y}\frac{dy}{dt}~dt = \frac{dz}{dt}~dt, \end{equation*}
which is the Chain Rule in this particular context, as expressed in Equation (11.6.2).
This approach to understanding the change in z using the differential or linearization also has the added benefit that we have separated how much of the change in \(x\) is coming through each of the intermediate variables \(x\) and \(y\text{.}\) Let’s go back to the context of our Preview Activity 11.6.1 where \(P\) is the amount of the large particulate matter in the air as function of location \((x,y)\) and our location is changing according to time along the path \(\vr(t)=\langle x(t),y(t)\rangle \text{.}\) The chain rule will express the rate of change in \(P\) as a function of time as
\begin{equation*} \frac{dP}{dt} = \frac{\partial P}{\partial x} \frac{dx}{dt}+ \frac{\partial P}{\partial y} \frac{dy}{dt} \end{equation*}
Because the amount of large particulate matter in terms of the \(x\) coordinate is governed by tree pollen, the term \(\frac{\partial P}{\partial x} \frac{dx}{dt}\) describes the rate of change in the large particulate matter per unit time along our drive due to tree pollen. Similarly, the amount of large particulate matter in terms of the \(y\) coordinate is governed by industrial pollution, so the second term, \(\frac{\partial P}{\partial y} \frac{dy}{dt}\text{,}\) describes the rate of change in the large particulate matter per unit time along our drive due to industrial pollution. Because our function is locally linear, the total rate of change in large particulate matter is the sum of these individual rates of change.

Activity 11.6.2.
In the following questions, we apply the Chain Rule in several different contexts.
(a)
Suppose that we have a function \(z\) defined by \(z(x,y) = x^2+xy^3\text{.}\) In addition, suppose that \(x\) and \(y\) are restricted to points that move around the plane by following a circle of radius \(2\) centered at the origin that is parameterized by
\begin{equation*} x(t) = 2\cos(t), \ \mbox{ and } \ y(t) = 2\sin(t) \end{equation*}
  1. Use the Chain Rule to find the resulting instantaneous rate of change \(\frac{dz}{dt}\text{.}\)
  2. Substitute \(x(t)\) for \(x\) and \(y(t)\) for \(y\) in the rule for \(z\) to write \(z\) in terms of \(t\) and calculate \(\frac{dz}{dt}\) directly.
  3. Write a couple of sentences comparing your answers to parts i. and ii.
(b)
Suppose that the temperature on a metal plate is given by the function \(T\) with
\begin{equation*} T(x,y) = 100-(x^2 + 4y^2) \end{equation*}
where the temperature is measured in degrees Fahrenheit and \(x\) and \(y\) are each measured in feet.
  1. Find \(T_x\) and \(T_y\text{.}\) What are the units on these partial derivatives?
  2. Suppose an ant is walking along the \(x\)-axis at the rate of 2 feet per minute toward the origin. When the ant is at the point \((2,0)\text{,}\) what is the instantaneous rate of change in the temperature \(dT/dt\) that the ant experiences. Include units on your response.
  3. Suppose instead that the ant walks along an ellipse with \(x = 6\cos(t)\) and \(y = 3\sin(t)\text{,}\) where \(t\) is measured in minutes. Find \(\frac{dT}{dt}\) at \(t = \pi/6\text{,}\) \(t=\pi/4\text{,}\) and \(t = \pi/3\text{.}\) What does this seem to tell you about the path along which the ant is walking?
(c)
Suppose that you are walking along a surface whose elevation is given by a function \(f\text{.}\) Furthermore, suppose that if you consider how your location corresponds to points in the \(xy\)-plane, you know that when you pass the point \((2,1)\text{,}\) your velocity vector is \(\vv=\langle -1,2\rangle\text{.}\) If some contours of \(f\) are as shown in Figure 11.6.2, estimate the rate of change \(df/dt\) when you pass through \((2,1)\text{.}\)
Figure 11.6.2. Some contours of \(f\text{.}\)
The previous activity and discussion have all dealt with a multivariable function where each of the inputs of the function depend on a separate variable, \(t\text{.}\) The next example generalizes the concept of the chain rule for multivariable functions with more than two inputs as well as the situation that the input variables depend on more than one separate variable.

Example 11.6.3.

Let \(C\) be the cost to manufacture flibertygibbits. The cost to produce flibertygibbits depends on the costs of two parts, which we will call \(p\) and \(q\text{,}\) and the cost of labor to construct the flibertygibbits, which we will call \(L\text{.}\) So \(C\) is a multivariable function of \(p\text{,}\) \(q\text{,}\) and \(L\text{,}\) which we will denote \(C(p,q,L)\text{.}\) The part costs (\(p\) and \(q\)) and the labor costs each depend on \(s\text{,}\) the distance to nearby suppliers and cities, and \(t\text{,}\) time. This means that \(p\text{,}\) \(q\text{,}\) and \(L\) are themselves multivariable functions of \(s\) and \(t\text{;}\) in other words we need to consider \(p(s,t)\text{,}\) \(q(s,t)\text{,}\) and \(L(s,t)\text{.}\)
All of these dependencies mean that \(C\) is dependent on \(s\) and \(t\) through composition with \(p(s,t)\text{,}\) \(q(s,t)\text{,}\) and \(L(s,t)\text{.}\) So we can ask how the cost to produce flibertygibbits depends on changes in \(s\) or \(t\text{,}\) which we can use the Chain Rule to describe. We can look at the differential of \(C\) in particular to see how these dependencies will affect each other. The differential
\begin{equation*} dC = \frac{\partial C}{\partial p} dp + \frac{\partial C}{\partial q} dq + \frac{\partial C}{\partial L} dL \end{equation*}
describes how a change in \(C\) will depend on changes in each variable \(p\text{,}\) \(q\text{,}\) and \(L\text{.}\)
We can look at how this change in \(C\) will depend on either \(s\) or \(t\) by taking the appropriate partial derivatives with respect to \(s\) or \(t\text{.}\) In particular, we get
\begin{equation*} \frac{\partial C}{\partial s} = \frac{\partial C}{\partial p} \frac{\partial p}{\partial s} + \frac{\partial C}{\partial q} \frac{\partial q}{\partial s} + \frac{\partial C}{\partial L} \frac{\partial L}{\partial s} \end{equation*}
and
\begin{equation*} \frac{\partial C}{\partial t} = \frac{\partial C}{\partial p} \frac{\partial p}{\partial t} + \frac{\partial C}{\partial q} \frac{\partial q}{\partial t} + \frac{\partial C}{\partial L} \frac{\partial L}{\partial t} \end{equation*}
These chain rule expressions have the same separation of intermediate variables as in our earlier discussion. For instance, the term \(\frac{\partial C}{\partial q} \frac{\partial q}{\partial s} \) describes the rate of change in the cost of making flibertygibbits per unit of distance from other cities that depends on the cost of part \(q\text{.}\) The term \(\frac{\partial C}{\partial L} \frac{\partial L}{\partial t} \) describes the rate of change in the cost of making flibertygibbits per unit time that depends on the cost of labor \(L\text{.}\) In both cases, the partial derivatives of \(C\) with respect to \(s\) and \(t\) are a linear combination of the changes through each of the intermediate variables \(p\text{,}\) \(q\text{,}\) and \(L\text{.}\)
We close this section with another statement on what linearity means in a bigger conceptual way. Every version of the chain rule we have looked at (including the chain rule from single variable calculus) states that the change in our output variable is a linear combination of the instantaneous rate of change in terms of each variable and the step size in that variable. The chain rule is simply an algebraic statement of this linear combination. As we said at the end of the previous section, if we are using a locally linear function then the instantaneous rate of change of the function will be the same as the corresponding measurement of change along the linearization/tangent plane.

Subsection 11.6.3 Summary

  • The Chain Rule is a tool for differentiating a composite for functions. In its simplest form, it says that if \(f(x,y)\) is a function of two variables and \(x(t)\) and \(y(t)\) depend on \(t\text{,}\) then
    \begin{equation*} \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}. \end{equation*}
  • The Chain Rule generalizes to functions of more than two variables and functions with more than one independent variable. If \(g(x,y,z)\) is a function of three variables and each of \(x\text{,}\) \(y\text{,}\) and \(z\) depend on \(s\) and \(t\text{,}\) then
    \begin{equation*} \frac{\partial g}{\partial s} = \frac{\partial g}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial g}{\partial y} \frac{\partial y}{\partial s} + \frac{\partial g}{\partial z} \frac{\partial a}{\partial s} \end{equation*}
    and
    \begin{equation*} \frac{\partial g}{\partial t} = \frac{\partial g}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial g}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial g}{\partial z} \frac{\partial z}{\partial t} \end{equation*}

Exercises 11.6.4 Exercises

1.

Use the chain rule to find \(\frac{dz}{dt}\text{,}\) where
\begin{equation*} z = x^2 y + x y^2,\quad x = -1+ t^7,\quad y = 2 - t^6 \end{equation*}
First the pieces:
\(\frac{\partial z}{\partial x} =\)
\(\frac{\partial z}{\partial y} =\)
\(\frac{dx}{dt} =\)
\(\frac{dy}{dt} =\)
End result (in terms of just \(t\)):
\(\frac{dz}{dt} =\)

2.

Use the chain rule to find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\text{,}\) where
\begin{equation*} z = e^{xy}\tan y, x = 3 s + 3 t, y = \frac{2 s}{4 t} \end{equation*}
First the pieces:
\(\frac{\partial z}{\partial x} =\) \(\frac{\partial z}{\partial y} =\)
\(\frac{\partial x}{\partial s} =\) \(\frac{\partial x}{\partial t} =\)
\(\frac{\partial y}{\partial s} =\) \(\frac{\partial y}{\partial t} =\)
And putting it all together:
\(\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\) and \(\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}\)

3.

Suppose \(\displaystyle w = \frac{x}{y} + \frac{y}{z}\text{,}\) where
\(x = e^{4t},\ y = 2 + \sin \left( t \right)\text{,}\) and \(z = 2 + \cos \left( 5t \right)\text{.}\)
A) Use the chain rule to find \(\frac{dw}{dt}\) as a function of x, y, z, and t. Do not rewrite x, y, and z in terms of t, and do not rewrite \(e^{4t}\) as x.
\(\frac{dw}{dt}\) =
Note: You may want to use exp() for the exponential function. Your answer should be an expression in x, y, z, and t; e.g. “3x - 4y”
B) Use part A to evaluate \(\frac{dw}{dt}\) when \(t = 0\text{.}\)

4.

If \(z = \left(x+y\right)e^{y}\) and \(x = u^{2}+v^{2}\) and \(y = u^{2}-v^{2}\text{,}\) find the following partial derivatives using the chain rule. Enter your answers as functions of \(u\) and \(v\text{.}\)
\(\displaystyle \frac{\partial z}{\partial u} =\)
\(\displaystyle \frac{\partial z}{\partial v} =\)

5.

If
\begin{equation*} z = \cos\!\left(x^{2}+y^{2}\right),\qquad x = v\cos\!\left(u\right),\qquad y = v\sin\!\left(u\right), \end{equation*}
find \(\partial z/\partial u\) and \(\partial z/\partial v\text{.}\) The variables are restricted to domains on which the functions are defined.
\(\partial z/\partial u =\)
\(\partial z/\partial v =\)

6.

Let \(z=g(u,v,w,x,y)\) and \(u(r,s),v(r,s),w(r,s),x(r,s),y(r,s)\text{.}\) How many terms are there in the expression for \(\partial z/\partial r\text{?}\)
terms

7.

Let \(W(s,t) = F(u(s,t), v(s,t))\) where
\begin{equation*} u(1,0) = -8, u_s(1,0) = 4, u_t(1,0) = -5 \end{equation*}
\begin{equation*} v(1,0) = -9, v_s(1,0) = -9, v_t(1,0) = 9 \end{equation*}
\begin{equation*} F_u(-8, -9) = -4, F_v(-8, -9) = -2 \end{equation*}
\(W_s(1,0) =\) \(W_t(1,0) =\)

8.

The radius of a right circular cone is increasing at a rate of 2 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 40 inches?
cubic inches per second

9.

In a simple electric circuit, Ohm’s law states that \(V = IR\text{,}\) where V is the voltage in volts, I is the current in amperes, and R is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.02 volts per second and, as the resistor heats up, the resistance is increasing at 0.01 ohms per second. When the resistance is 100 ohms and the current is 0.02 amperes, at what rate is the current changing?
amperes per second

10.

Suppose \(z = x^{2} \sin y\text{,}\) \(x = 2 s^{2} + 2 t^{2}\text{,}\) \(y = -4 s t\text{.}\)
A. Use the chain rule to find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) as functions of x, y, s and t.
\(\frac{\partial z}{\partial s} =\)
\(\frac{\partial z}{\partial t} =\)
B. Find the numerical values of \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) when \(\left( s , t \right) = \left( 3 , 1 \right)\text{.}\)
\(\frac{\partial z}{\partial s} \left( 3 , 1 \right) =\)
\(\frac{\partial z}{\partial t} \left( 3 , 1 \right) =\)

11.

Find the indicated derivative. In each case, state the version of the Chain Rule that you are using.
  1. \(\frac{df}{dt}\text{,}\) if \(f(x,y) = 2x^2y\text{,}\) \(x=\cos(t)\text{,}\) and \(y=\ln(t)\text{.}\)
  2. \(\frac{\partial f}{\partial w}\text{,}\) if \(f(x,y) = 2x^2y\text{,}\) \(x=w+z^2\text{,}\) and \(y=\frac{2z+1}{w}\)
  3. \(\frac{\partial f}{\partial v}\text{,}\) if \(f(x,y,z) = 2x^2y+z^3\text{,}\) \(x=u-v+2w\text{,}\) \(y=w2^v-u^3\text{,}\) and \(z = u^2-v\)

12.

Let \(z = u^2 - v^2\) and suppose that
\begin{align*} u \amp = e^x\cos(y)\\ v \amp = e^x\sin(y) \end{align*}
  1. Find the values of \(u\) and \(v\) that correspond to \(x=0\) and \(y=2\pi/3\text{.}\)
  2. Use the Chain Rule to find the general partial derivatives
    \begin{equation*} \frac{\partial z}{\partial x} \ \mbox{and} \ \frac{\partial z}{\partial y} \end{equation*}
    and then determine both \(\frac{\partial z}{\partial x}\restrict{(0, \frac{2\pi}{3})}\) and \(\frac{\partial z}{\partial y}\restrict{(0, \frac{2\pi}{3})}\text{.}\)

13.

Suppose that \(T = x^2 + y^2 - 2z\) where
\begin{align*} x \amp = \rho\sin(\phi)\cos(\theta)\\ y \amp = \rho\sin(\phi)\sin(\theta)\\ z \amp = \rho\cos(\phi) \end{align*}
  1. Construct a tree diagram representing the dependencies among the variables.
  2. Apply the chain rule to find the partial derivatives
    \begin{equation*} \frac{\partial T}{\partial\rho}, \frac{\partial T}{\partial\phi}, \ \mbox{and} \ \frac{\partial T}{\partial\theta}. \ \end{equation*}

14.

Suppose that the temperature on a metal plate is given by the function \(T\) with
\begin{equation*} T(x,y) = 100-(x^2 + 4y^2), \end{equation*}
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 \((5,6)\text{.}\)
  1. Find parametric equations \((x(t),y(t))\) for the ant’s coordinates as it walks the line from \((1,4)\) to \((5,6)\text{.}\)
  2. What can you say about \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) for every value of \(t\text{?}\)
  3. 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 \((5,6)\text{,}\) using your parametric equations for \(x\) and \(y\text{.}\) Include units on your answer.

15.

There are several proposed formulas to approximate the surface area of the human body. One model
 2 
DuBois D, DuBois DF. A formula to estimate the approximate surface area if height and weight be known. Arch Int Med 1916;17:863-71.
uses the formula
\begin{equation*} A(h,w) = 0.0072 h^{0.725}w^{0.425}, \end{equation*}
where \(A\) is the surface area in square meters, \(h\) is the height in centimeters, and \(w\) is the weight in kilograms.
Since a person’s height \(h\) and weight \(w\) change over time, \(h\) and \(w\) are functions of time \(t\text{.}\) Let us think about what is happening to a child whose height is \(60\) centimeters and weight is \(9\) kilograms. Suppose, furthermore, that \(h\) is increasing at an instantaneous rate of 20 centimeters per year and \(w\) is increasing at an instantaneous rate of \(5\) kg per year.
Determine the instantaneous rate at which the child’s surface area is changing at this point in time.

16.

Let \(z = f(x,y) = 50 - (x+1)^2 - (y+3)^2\) and \(z = h(x,y) = 24 - 2x - 6y\text{.}\)
Suppose a person is walking on the surface \(z = f(x,y)\) in such a way that she walks the curve which is the intersection of \(f\) and \(h\text{.}\)
  1. Show that \(x(t) = 4 \cos(t)\) and \(y(t) = 4 \sin(t)\) is a parameterization of the “shadow” in the \(xy\)-plane of the curve that is the intersection of the graphs of \(f\) and \(h\text{.}\)
  2. Use the parameterization from part (a) to find the instantaneous rate at which her height is changing with respect to time at the instant \(t = 2\pi/3\text{.}\)

17.

The voltage \(V\) (in volts) across a circuit is given by Ohm’s Law: \(V = IR\text{,}\) where \(I\) is the current (in amps) in the circuit and \(R\) is the resistance (in ohms). Suppose we connect two resistors with resistances \(R_1\) and \(R_2\) in parallel as shown in Figure 11.6.4. The total resistance \(R\) in the circuit is then given by
\begin{equation*} \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}. \end{equation*}
  1. Assume that the current, \(I\text{,}\) and the resistances, \(R_1\) and \(R_2\text{,}\) are changing over time, \(t\text{.}\) Use the Chain Rule to write a formula for \(\frac{dV}{dt}\text{.}\)
  2. Suppose that, at some particular point in time, we measure the current to be 3 amps and that the current is increasing at \(\frac{1}{10}\) amps per second, while resistance \(R_1\) is 2 ohms and decreasing at the rate of 0.2 ohms per second and \(R_2\) is 1 ohm and increasing at the rate of \(0.5\) ohms per second. At what rate is the voltage changing at this point in time?
Figure 11.6.4. Resistors in parallel.