Multivariable calculus: Exercises

12 Chain rule

Exercise 1: Practice with matrix multiplication

In this exercise, you’ll get some warm-up practice with matrix multiplication to prepare for the chain rule.

Compute the product \[ \begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & -1 \\ 0 & 4 \end{bmatrix}. \]
Compute the product \[ \begin{bmatrix} 3 & -1 \\ 2 & 0 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 & -1 \\ 1 & 3 & 2 & 1 \end{bmatrix}. \]
Compute the product \[ \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 3 \\ -2 \end{bmatrix}. \]
Compute the product \[ \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}. \]
Compute the product \[ \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 0 \\ -1 & 4 \end{bmatrix}. \]

Solutions.

Answer: \[ \begin{bmatrix} 4 & 11 \end{bmatrix} \]
Answer: \[ \begin{bmatrix} 5 & 0 & -2 & -4 \\ 4 & 2 & 0 & -2 \\ 6 & 13 & 8 & 3 \end{bmatrix} \]
Answer: \[ \begin{bmatrix} -1 \\ 2 \end{bmatrix} \]
Answer: \[ \begin{bmatrix} 2x - y + 3z \end{bmatrix} \]
Answer: \[ \begin{bmatrix} \cos{\theta} - 3\sin{\theta} & 2\cos{\theta} \\ 3 & 0 \end{bmatrix} \]

Exercise 2: Practice with the chain rule

Now that you’ve seen how the chain rule works by computing both sides, let’s practice using it directly to compute derivatives of compositions.

Consider the functions \[ \mathbb{R}^2 \xrightarrow{f} \mathbb{R}^3 \xrightarrow{g} \mathbb{R} \] where \[ f(x,y) = (x^2, xy, y^2), \quad g(u,v,w) = u + 2v + 3w. \] Compute the derivative \((g\circ f)'(x,y)\) using the chain rule.
Consider the functions \[ \mathbb{R} \xrightarrow{h} \mathbb{R}^2 \xrightarrow{k} \mathbb{R}^3 \] where \[ h(t) = (\cos{t}, \sin{t}), \quad k(u,v) = (u^2v, u-v, v^3). \] Compute the derivative \((k\circ h)'(t)\) using the chain rule.
Consider the functions \[ \mathbb{R}^3 \xrightarrow{p} \mathbb{R}^2 \xrightarrow{q} \mathbb{R}^2 \] where \[ p(x,y,z) = (x+y+z, xyz), \quad q(r,s) = (r^2s, e^{rs}). \] Compute the derivative \((q\circ p)'(1,1,1)\) using the chain rule.
Consider the functions \[ \mathbb{R}^2 \xrightarrow{f} \mathbb{R}^2 \xrightarrow{g} \mathbb{R}^3 \] where \[ f(x,y) = (e^x\cos{y}, e^x\sin{y}), \quad g(u,v) = (u, v, u^2+v^2). \] Compute the derivative \((g\circ f)'(0, \pi/4)\) using the chain rule.
Consider the function \[ \mathbb{R}^4 \xrightarrow{h} \mathbb{R}^2 \xrightarrow{k} \mathbb{R} \] where \[ h(x,y,z,w) = (x^2 + w^2, y^2 + z^2), \quad k(u,v) = uv. \] Compute the derivative \((k\circ h)'(1, -1, 0, 1)\) using the chain rule.
Consider the functions \[ \mathbb{R} \xrightarrow{\alpha} \mathbb{R}^3 \xrightarrow{\beta} \mathbb{R} \] where \[ \alpha(t) = (t, t^2, t^3), \quad \beta(x,y,z) = x^2y + yz^2. \] Compute the derivative \((\beta\circ \alpha)'(2)\) using the chain rule.

Solutions.

Answer: \[ (g\circ f)'(x,y) = \begin{bmatrix} 2x + 2y & 2x + 6y \end{bmatrix} \]
Answer: \[ (k\circ h)'(t) = \begin{bmatrix} -2\cos{t}\sin^2{t} + \cos^3{t} \\ -\sin{t} - \cos{t} \\ 3\sin^2{t}\cos{t} \end{bmatrix} \]
Answer: \[ (q\circ p)'(1,1,1) = \begin{bmatrix} 15 & 15 & 15 \\ 4e^3 & 4e^3 & 4e^3 \end{bmatrix} \]
Answer: \[ \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 2 & 0 \end{bmatrix} \]
Answer: \[ \begin{bmatrix} 2 & -4 & 0 & 2 \end{bmatrix} \]
Answer: \(1056\).

Exercise 3: The chain rule and partial derivatives

Recall that when we write the chain rule in terms of partial derivatives, it takes the form \[ \frac{\partial (g\circ f)_i}{\partial x_j}(\mathbf{x}) = \sum_{p=1}^m \frac{\partial g_i}{\partial u_p}(f(\mathbf{x}))\frac{\partial f_p}{\partial x_j}(\mathbf{x}), \] where the \(u_p\) are the variables in the domain of \(g\). Very often, we use a shorthand notation that omits the evaluations and lets the \(u_p\) play a double role as both coordinate functions of \(f\) and variables in the domain of \(g\).

Let \(z = u^3 + uv^2\), where \(u = x^2y\) and \(v = xy^2\). Compute \(\displaystyle\frac{\partial z}{\partial x}\) and \(\displaystyle\frac{\partial z}{\partial y}\) using the chain rule.
Let \(w(s,t) = s^2\cos{t} + e^{st}\), where \(s = r + 2\theta\) and \(t = r\theta\). Compute \(\displaystyle\frac{\partial w}{\partial r}\) and \(\displaystyle\frac{\partial w}{\partial \theta}\) using the chain rule.
Let \(f(u,v,w) = (u^2v, vw^2, u+v+w)\), where \(u = x+y\), \(v = xy\), and \(w = x-y\). Compute all six partial derivatives \[ \frac{\partial f_1}{\partial x}, \quad \frac{\partial f_1}{\partial y}, \quad \frac{\partial f_2}{\partial x}, \quad \frac{\partial f_2}{\partial y}, \quad \frac{\partial f_3}{\partial x}, \quad \frac{\partial f_3}{\partial y} \] using the chain rule.

Solutions.

Answers: \[ \begin{align*} \frac{\partial z}{\partial x} &= 6x^5y^3 + 4x^3y^5 \\ \frac{\partial z}{\partial y} &= 3x^6y^2 + 5x^4y^4 \end{align*} \]
Answers: \[ \begin{align*} \frac{\partial w}{\partial r} &= 2 (2 \theta +r) \cos (\theta r)-\theta (2 \theta +r)^2 \sin (\theta r) \\ \frac{\partial w}{\partial \theta} &= 4 (2 \theta +r) \cos (\theta r)-r (2 \theta +r)^2 \sin (\theta r) \end{align*} \]
Answers: \[ \begin{align*} \frac{\partial f_1}{\partial x} &= 3x^2y + 4xy^2 + y^3 \\ \frac{\partial f_1}{\partial y} &= x^3 + 4x^2y + 3xy^2 \\ \frac{\partial f_2}{\partial x} &= 3x^2y - 4xy^2 + y^3 \\ \frac{\partial f_2}{\partial y} &= x^3 - 4x^2y + 3xy^2 \\ \frac{\partial f_3}{\partial x} &= 2 + y \\ \frac{\partial f_3}{\partial y} &= x \end{align*} \]

Exercise 4: Application problems

The chain rule is particularly useful for solving related rates problems, where we need to understand how different quantities change with respect to time.

A rectangular box has dimensions that are all changing with time. At a particular moment, the length is \(5\) m and increasing at \(0.3\) m/s, the width is \(4\) m and decreasing at \(0.2\) m/s, and the height is \(3\) m and increasing at \(0.5\) m/s. Is the volume of the box increasing or decreasing at this moment, and at what rate?
The position of a particle at time \(t\) is given by \(x=t^2\) and \(y=2t-1\). The temperature at a point \((x,y)\) in the plane is given by \(T(x,y) = x^2 + xy + y^2\). Find the rate of change of temperature experienced by the particle at time \(t=2\).

Solutions.

The volume of a rectangular box is \(V = \ell wh\). We are given that \(\ell = 5\) m, \(w = 4\) m, \(h = 3\) m, \(\displaystyle\frac{d\ell}{dt} = 0.3\) m/s, \(\displaystyle\frac{dw}{dt} = -0.2\) m/s, and \(\displaystyle\frac{dh}{dt} = 0.5\) m/s. Using the chain rule: \[ \frac{dV}{dt} = \frac{\partial V}{\partial \ell}\frac{d\ell}{dt} + \frac{\partial V}{\partial w}\frac{dw}{dt} + \frac{\partial V}{\partial h}\frac{dh}{dt} = (wh)(0.3) + (\ell h)(-0.2) + (\ell w)(0.5). \] Substituting the given values: \[ \frac{dV}{dt} = (4)(3)(0.3) + (5)(3)(-0.2) + (5)(4)(0.5) = 3.6 - 3 + 10 = 10.6 \text{ m}^3\text{/s}. \]
The rate of change of temperature is \(\displaystyle\frac{dT}{dt}\). Using the chain rule: \[ \frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}. \] We have: \[ \frac{\partial T}{\partial x} = 2x + y, \quad \frac{\partial T}{\partial y} = x + 2y, \quad \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 2. \] At \(t = 2\), we have \((x,y) = (4, 3)\). Therefore: \[ \frac{dT}{dt}\bigg|_{t=2} = (2(4) + 3)(2(2)) + (4 + 2(3))(2) = (11)(4) + (10)(2) = 44 + 20 = 64 \text{ degrees per second}. \]