Multivariable calculus: Exercises

13 Second derivatives, part 1

Exercise 1: What is a second derivative?

Let \(f:\mathbb{R}^n \to \mathbb{R}\) be a function whose second derivative exists.

1. State the precise, mathematical definition of the second derivative \(\mathbf{h}^\top f''(\mathbf{v})\mathbf{h}\) at a point \(\mathbf{v}\) in \(\mathbb{R}^n\) in the direction of a step vector \(\mathbf{h}\). What type of object is \(f''(\mathbf{v})\)?

2. Given a position vector \(\mathbf{v}\) and a unit step vector \(\mathbf{h}\), what does the number \(\mathbf{h}^\top f''(\mathbf{v})\mathbf{h}\) measure? What does it mean if this number is positive? Negative? Briefly explain in your own words.

NoteSolutions.

All the solutions are in your class notes and lecture slides. Look for them there.

Exercise 2: Practice computing second derivatives from the definition

Compute the second derivatives of the following functions using the definition of the second derivative.

1. Given \(f:\mathbb{R}^2 \to \mathbb{R}\), \(f(x,y) = x^2 - y^2\), compute \(f''(x,y)\).

2. Given \(g:\mathbb{R}^3 \to \mathbb{R}\), \(g(x,y,z) = xy + z^3\), compute \(g''(x,y,z)\).

3. Given \(h:\mathbb{R}^2 \to \mathbb{R}\), \(h(x,y) = e^{x^2+y}\), compute \(h''(1,1)\).

NoteSolutions.

1. Answer: \[ f''(x,y) = \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix} \]

2. Answer: \[ g''(x,y,z) = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6z \end{bmatrix} \]

3. Answer: \[ h''(1,1) = \begin{bmatrix} 6e^2 & 2e^2 \\ 2e^2 & e^2 \end{bmatrix} \]

Exercise 3: Practice computing second derivatives using partial derivatives

Compute the second derivatives of the following functions using partial derivatives.

1. Given \(f:\mathbb{R}^3 \to \mathbb{R}\), \(f(x,y,z) = e^z\cos{x} + xy\), compute \(f''(x,y,z)\).

2. Given \(g:\mathbb{R}^2 \to \mathbb{R}\), \(g(x,y) = 4x^2 + e^xx^2\), compute \(g''(x,y)\).

3. Given \(h:\mathbb{R}^4 \to \mathbb{R}\), \(h(x,y,z,w) = we^{x^2+y} + z^2\), compute \(h''(x,y,z,w)\).

Now find the curvatures of the functions at the following points in the direction of the given step vectors.

4. At \((0,1,0)\) in the direction of \(\mathbf{h}^\top = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}\) for \(f\).

5. At \((0,0)\) in the direction of \(\mathbf{h}^\top = \begin{bmatrix} 1 & 1 \end{bmatrix}\) for \(g\).

6. At \((1,0,-1,0)\) in the direction of \(\mathbf{h}^\top = \begin{bmatrix} 1 & 1 & 0 & 0 \end{bmatrix}\) for \(h\).

NoteSolutions.

1. Answer: \[ f''(x,y,z) = \begin{bmatrix} -e^z\cos{x} & 1 & -e^z\sin{x} \\ 1 & 0 & 0 \\ -e^z\sin{x} & 0 & e^z\cos{x} \end{bmatrix} \]

2. Answer: \[ g''(x,y) = \begin{bmatrix} 8 + 2e^x + 4e^xx+ e^x x^2 & 0 \\ 0 & 0 \end{bmatrix} \]

3. Answer: \[ h''(x,y,z,w) = \begin{bmatrix} 2we^{x^2+y} + 4x^2we^{x^2+y} & 2xwe^{x^2+y} & 0 & 2e^{x^2+y} \\ 2xwe^{x^2+y} & we^{x^2+y} & 0 & e^{x^2+y} \\ 0 & 0 & 2 & 0 \\ 2xe^{x^2+y} & e^{x^2+y} & 0 & 0 \end{bmatrix} \]

4. Answer: \(-1\)

5. Answer: \(5\)

6. Answer: \(0\)

Exercise 4: Visualizing curvature

Consider the graph of the function \(f(x,y) = x^2 - y^2+4xy^2\).

1. Find a unit step vector \(\mathbf{h}\) at the point \((0,0)\) for which the curvature in that direction is positive. Confirm your answer by computing the curvature in that direction.

2. Now find a unit step vector \(\mathbf{h}\) at the point \((0,0)\) for which the curvature in that direction is negative. Confirm your answer by computing the curvature in that direction.

NoteSolutions.

The only part I will give you is the second derivative:

\[ f''(x,y) = \begin{bmatrix} 2 & 8y \\ 8y & -2 + 8x \end{bmatrix} \]

It’s up to you to find the two unit step vectors.