Multivariable calculus: Exercises

11 Partial derivatives, part 3

Exercise 1: Computing derivatives

Recall from the class that the derivative of a function \(f: \mathbb{R}^n \to \mathbb{R}^m\) may be computed as the \(m\times n\) matrix

\[ f'(\mathbf{v}) = \begin{bmatrix} \displaystyle\frac{\partial f_1}{\partial x_1}(\mathbf{v}) & \displaystyle\frac{\partial f_1}{\partial x_2}(\mathbf{v}) & \cdots & \displaystyle\frac{\partial f_1}{\partial x_n}(\mathbf{v}) \\ \displaystyle\frac{\partial f_2}{\partial x_1}(\mathbf{v}) & \displaystyle\frac{\partial f_2}{\partial x_2}(\mathbf{v}) & \cdots & \displaystyle\frac{\partial f_2}{\partial x_n}(\mathbf{v}) \\ \vdots & \vdots & \ddots & \vdots \\ \displaystyle\frac{\partial f_m}{\partial x_1}(\mathbf{v}) & \displaystyle\frac{\partial f_m}{\partial x_2}(\mathbf{v}) & \cdots & \displaystyle\frac{\partial f_m}{\partial x_n}(\mathbf{v}) \end{bmatrix}, \]

where \(f_1,f_2,\ldots,f_m\) are the coordinate functions of \(f\) and I’ve put the variables into a vector to shorten the notation:

\[ \mathbf{v} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}. \]

Compute the derivatives of the following functions:

1. \[f:\mathbb{R}^2 \to \mathbb{R}^3, \quad f(x,y) = (x^2+y^2, xy^2, x+y)\]
2. \[g:\mathbb{R}^3 \to \mathbb{R}^2, \quad g(x,y,z) = \left(\frac{3xz}{x+y}, x^2+y^2+z^2\right)\]
3. \[h:\mathbb{R}^3 \to \mathbb{R}^3, \quad h(x,y,z) = (\sin(x+2y+3z), e^{x+y+z}, x^2+y^2+z^2)\]
4. \[k:\mathbb{R} \to \mathbb{R}^4, \quad k(t) = (t, t^2, t^3, t^4)\]
5. \[u:\mathbb{R}^3 \to \mathbb{R}^4, \quad u(x,y,z) = (x\cos{y}, x\sin{y}, y\cos{z}, y\sin{z})\]
6. \[v:\mathbb{R}^4 \to \mathbb{R}^3, \quad v(x,y,z,w) = (x+y+z+w, xyzw, x^2+y^2+z^2+w^2)\]

NoteSolutions.

1. The derivative is \[ f'(x,y) = \begin{bmatrix} 2x & 2y \\ y^2 & 2xy \\ 1 & 1 \end{bmatrix}. \]

2. The derivative is \[ g'(x,y,z) = \begin{bmatrix} \displaystyle\frac{3yz}{(x+y)^2} & \displaystyle\frac{-3xz}{(x+y)^2} & \displaystyle\frac{3x}{x+y} \\ 2x & 2y & 2z \end{bmatrix}. \]

3. The derivative is \[ h'(x,y,z) = \begin{bmatrix} \cos(x+2y+3z) & 2\cos(x+2y+3z) & 3\cos(x+2y+3z) \\ e^{x+y+z} & e^{x+y+z} & e^{x+y+z} \\ 2x & 2y & 2z \end{bmatrix}. \]

4. The derivative is \[ k'(t) = \begin{bmatrix} 1 \\ 2t \\ 3t^2 \\ 4t^3 \end{bmatrix}. \]

5. The derivative is \[ u'(x,y,z) = \begin{bmatrix} \cos{y} & -x\sin{y} & 0 \\ \sin{y} & x\cos{y} & 0 \\ 0 & \cos{z} & -y\sin{z} \\ 0 & \sin{z} & y\cos{z} \end{bmatrix}. \]

6. The derivative is \[ v'(x,y,z,w) = \begin{bmatrix} 1 & 1 & 1 & 1 \\ yzw & xzw & xyw & xyz \\ 2x & 2y & 2z & 2w \end{bmatrix}. \]

Exercise 2: Tangent approximations

Recall that the tangent approximation of a function \(f: \mathbb{R}^n \to \mathbb{R}^m\) at a point \(\mathbf{v}_0\) is given by \[ L(\mathbf{v}) = f'(\mathbf{v}_0)(\mathbf{v}-\mathbf{v}_0) + f(\mathbf{v}_0), \]

where \(\mathbf{v}\) is a variable position vector.

Consider the function

\[ f: \mathbb{R}^3 \to \mathbb{R}^2, \quad f(x,y,z) = (x^2+y^2, xy+yz-2xz). \]

1. Compute the derivative \(f'(x,y,z)\).
2. Compute the tangent approximation of \(f\) at the point \((1,0,-1)\).
3. Find all points \(\mathbf{v_0}\) where the tangent approximation of \(f\) is constant.
4. At the point that you found in part (c), what does a constant tangent approximation tell you about the function \(f\) near that point?

NoteSolutions.

1. The derivative is \[ f'(x,y,z) = \begin{bmatrix} 2x & 2y & 0 \\ y-2z & x+z & y-2x \end{bmatrix}. \]

2. The tangent approximation at \((1,0,-1)\) is \[ L(x,y,z) = \begin{bmatrix} 2 & 0 & 0 \\ 2 & 0 & -2 \end{bmatrix} \begin{bmatrix} x-1 \\ y \\ z+1 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2x-1 \\ 2x-2z - 2 \end{bmatrix} \]

3. Points \(\mathbf{v}_0\) at which the tangent approximation is constant are those for which the derivative \(f'(\mathbf{v}_0)\) is the zero matrix. Do you understand why? This is because if \(f'(\mathbf{v}_0)\) is the zero matrix, then \(L(\mathbf{v}) = f(\mathbf{v}_0)\) for all \(\mathbf{v}\), which is a constant function. So we need to solve the system of equations \[ \begin{bmatrix} 2x_0 & 2y_0 & 0 \\ y_0-2z_0 & x_0+z_0 & y_0-2x_0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \] But we have \(2x_0 = 0\) and \(2y_0 = 0\), which implies that \(x_0 = 0\) and \(y_0 = 0\). Then the second row of the matrix gives us the equations \[ y_0 - 2z_0 = 0, \quad x_0 + z_0 = 0, \quad y_0 - 2x_0 = 0. \] However, we know that \(x_0 = 0\) and \(y_0 = 0\), so these equations tell us \(z_0 = 0\). Therefore, the only point at which the tangent approximation is constant is \(\mathbf{v}_0 = (0,0,0)\).

4. We know that, for small step vectors \(\mathbf{h}\), we have \[ f(\mathbf{v}_0 + \mathbf{h}) - f(\mathbf{v}_0) \approx f'(\mathbf{v}_0)\mathbf{h}. \] But if \(f'(\mathbf{v}_0)\) is the zero matrix, then \(f(\mathbf{v}_0 + \mathbf{h}) - f(\mathbf{v}_0) \approx 0\), which means that \(f(\mathbf{v}_0 + \mathbf{h}) \approx f(\mathbf{v}_0)\) for small \(\mathbf{h}\). In other words, the function \(f\) is approximately constant near the point \(\mathbf{v}_0 = (0,0,0)\).

Exercise 3: Application problem

We’ve been focusing so much on the mechanics of computing derivatives that we haven’t had much time to discuss applications. Let’s fix that with a simple application problem.

A factory produces two products. The production outputs depend on labor \(L\) and capital \(K\) according to: \[ P(L,K) = (20L^{0.6}K^{0.4}, \, 15L^{0.5}K^{0.5}) \] where the coordinate functions \(P_1\) and \(P_2\) are the quantities of products 1 and 2 produced.

Currently, the factory employs \(L=100\) units of labor and \(K=50\) units of capital.

1. Find the derivative \(P'(100, 50)\).
2. If the factory increases labor by 2 units and capital by 1 unit, approximately how much does the production of each product increase?

NoteSolutions.

1. After rounding, the derivative is \[ P'(100,50) \approx \begin{bmatrix} 9.0943 & 12.1257 \\ 5.3033 & 10.6066 \\ \end{bmatrix}. \]

2. We have \[ P(102,51) - P(100,50) \approx P'(100,50) \begin{bmatrix} 2 \\ 1 \end{bmatrix} \approx \begin{bmatrix} 9.0943 & 12.1257 \\ 5.3033 & 10.6066 \\ \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 30.3143 \\ 21.2132 \\ \end{bmatrix}. \] Thus, the production of product 1 increases by approximately 30.31 units and the production of product 2 increases by approximately 21.21 units.