Multivariable calculus: Exercises

08 The definition of the derivative

Exercise 1: Computing derivatives from the definition

Recall the definition of the derivative of a function \(f: \mathbb{R}^n \to \mathbb{R}^m\) at a point \(\mathbf{v}_0\) given in class:

\[ f'(\mathbf{v}_0) \mathbf{h} = \lim_{\lambda \to 0} \frac{f(\mathbf{v}_0 + \lambda\mathbf{h}) - f(\mathbf{v}_0)}{\lambda}, \]

for all \(\mathbf{h}\) in \(\mathbb{R}^n\).

Using the above definition, compute the derivatives of the following functions.

  1. \(f: \mathbb{R}^2 \to \mathbb{R}\) given by \(f(x,y) = x^2 - y^2\). Compute \(f'(x,y)\).
  2. \(g: \mathbb{R}^2 \to \mathbb{R}\) given by \(g(x,y) = 4-x^2-4y^2\). Compute \(g'(x,y)\).
  3. \(q: \mathbb{R} \to \mathbb{R}^2\) given by \(q(t) = (t + 2t^2, -t^3)\). Compute \(q'(t)\).
  4. \(r: \mathbb{R} \to \mathbb{R}^3\) given by \(r(t) = (e^t, \sin{t}, \cos{t})\). Compute \(r'(t)\).
  5. \(u:\mathbb{R}^3 \to \mathbb{R}\) given by \(u(x,y,z) = x^2 + y^2 + z^2\). Compute \(u'(x,y,z)\).
  6. \(p: \mathbb{R}^2 \to \mathbb{R}^3\) given by \(p(x,y) = (x^2, 3y, y^2)\). Compute \(p'(x,y)\).

  1. \(f'(x,y) = \begin{bmatrix} 2x & -2y \end{bmatrix}\).

  2. \(g'(x,y) = \begin{bmatrix} -2x & -8y \end{bmatrix}\).

  3. \(q'(t) = \begin{bmatrix} 1 + 4t \\ -3t^2 \end{bmatrix}\).

  4. \(r'(t) = \begin{bmatrix} e^t \\ \cos{t} \\ -\sin{t} \end{bmatrix}\).

  5. \(u'(x,y,z) = \begin{bmatrix} 2x & 2y & 2z \end{bmatrix}\).

  6. \(p'(x,y) = \begin{bmatrix} 2x & 0 \\ 0 & 3 \\ 0 & 2y \end{bmatrix}\).

Exercise 2: Computing tangent approximations and spaces

Recall that the tangent space to a function \(f: \mathbb{R}^n \to \mathbb{R}^m\) at a point \(\mathbf{v}_0\) is the range of the tangent approximation function:

\[ L_{\mathbf{v}_0}(\mathbf{v}) = f'(\mathbf{v}_0)(\mathbf{v} - \mathbf{v}_0) + f(\mathbf{v}_0). \]

This exercise refers to the functions defined in Exercise 1.

  1. For \(f(x,y) = x^2 - y^2\), compute the tangent approximation at the point \((x_0,y_0)=(1,0)\). Compute the tangent plane at this point as well, and check your answer using Desmos.

  2. For \(g(x,y) = 4-x^2-4y^2\), compute the tangent approximation at the point \((x_0,y_0)=(1,1)\). Compute the tangent plane at this point as well, and check your answer using Desmos.

  3. For \(q(t) = (t + 2t^2, -t^3)\), compute the tangent approximation at the point \(t_0=1\). Compute the tangent line at this point as well, and check your answer using Desmos.

  4. For \(r(t) = (e^t, \sin{t}, \cos{t})\), compute the tangent approximation at the point \(t_0=0\). Compute the tangent line at this point as well, and check your answer using Desmos.

  5. For \(u(x,y,z) = x^2 + y^2 + z^2\), compute the tangent approximation at the point \((x_0,y_0,z_0)=(1,1,1)\).

  6. For \(p(x,y) = (x^2, 3y, y^2)\), compute the tangent approximation at the point \((x_0,y_0)=(1,1)\).

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

  2. The tangent approximation is \[ L(x,y) = \begin{bmatrix} -2 & -8 \end{bmatrix} \begin{bmatrix} x-1 \\ y-1 \end{bmatrix} - 1 = -2x - 8y + 9. \] The tangent plane is here.

  3. The tangent approximation is \[ L(t) = \begin{bmatrix} 5 \\ -3 \end{bmatrix} (t-1) + \begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 5t - 2 \\ -3t + 2 \end{bmatrix}. \] The tangent line is here.

  4. The tangent approximation is \[ L(t) = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} t + \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} t + 1 \\ t \\ 1 \end{bmatrix}. \] The tangent line is here.

  5. The tangent approximation is \[ L(x,y,z) = \begin{bmatrix} 2 & 2 & 2 \end{bmatrix} \begin{bmatrix} x-1 \\ y-1 \\ z-1 \end{bmatrix} + 3 = 2x + 2y + 2z - 3. \]

  6. The tangent approximation is \[ L(x,y) = \begin{bmatrix} 2 & 0 \\ 0 & 3 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} x-1 \\ y-1 \end{bmatrix} + \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2x - 1 \\ 3y \\ 2y - 1 \end{bmatrix}. \]

Exercise 3: Normal and unit vectors

For this exercise, you’ll need to remember the definition of normal vectors to planes, and unit vectors. Go back and look them up!

  1. In Exercise 2(a), you computed the tangent plane to the function \(f\) at the specified point. Compute an upward-pointing unit normal vector. (“Upward-pointing” means that the \(z\)-component of the normal vector should be positive.) Check your answer using Desmos.

  2. In Exercise 2(b), you computed the tangent plane to the function \(g\) at the specified point. Compute an upward-pointing unit normal vector. Check your answer using Desmos.

  3. In Exercise 2(c), you computed the tangent line to the function \(q\) at the specified points. Compute a positively-oriented unit tangent vector at the point. (“Tangent vector” means it lies in the tangent line, and “positively-oriented” means that the vector points in the direction that the curve is traced out as \(t\) increases.) Check your answer using Desmos.

  4. In Exercise 2(d), you computed the tangent line to the function \(r\) at the specified points. Compute a positively-oriented unit tangent vector at the point. Check your answer using Desmos.

  1. The tangent plane for \(f\) had equation \(z = 2x-1\). Since this is in point-slope form, as we learned in class, an upward-pointing normal vector is \[ \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}. \] To make this a unit vector, we divide by its length: \[ \mathbf{n} = \frac{1}{\sqrt{5}} \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -\frac{2}{\sqrt{5}} \\ 0 \\ \frac{1}{\sqrt{5}} \end{bmatrix}. \] The graph is here.

  2. The tangent plane for \(g\) had equation \(z = -2x - 8y + 9\). An upward-pointing normal vector is \[ \begin{bmatrix} 2 \\ 8 \\ 1 \end{bmatrix}. \] To make this a unit vector, we divide by its length: \[ \mathbf{n} = \frac{1}{\sqrt{69}} \begin{bmatrix} 2 \\ 8 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{\sqrt{69}} \\ \frac{8}{\sqrt{69}} \\ \frac{1}{\sqrt{69}} \end{bmatrix}. \] The graph is here.

  3. The tangent line for \(q\) is the image of the function \[ L(t) = \begin{bmatrix} 5 \\ -3 \end{bmatrix} (t-1) + \begin{bmatrix} 3 \\ -1 \end{bmatrix}. \] Thus, for all \(t\), the vector \[ L(t) - \begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 5 \\ -3 \end{bmatrix} (t-1) \] must lie in the tangent line. (Convince yourself of this by drawing a picture!) We may as well pick \(t=2\) to get a nonzero vector in the tangent line: \[ \begin{bmatrix} 5 \\ -3 \end{bmatrix}. \] This vector points in the direction that the curve is traced out as \(t\) increases, so it is a positively-oriented tangent vector. To make it a unit vector, we divide by its length: \[ \mathbf{t} = \frac{1}{\sqrt{34}} \begin{bmatrix} 5 \\ -3 \end{bmatrix} = \begin{bmatrix} \frac{5}{\sqrt{34}} \\ -\frac{3}{\sqrt{34}} \end{bmatrix}. \] The graph is here.

  4. The tangent line for \(r\) is the image of the function \[ L(t) = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} t + \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}. \] Thus, for all \(t\), the vector \[ L(t) - \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} t \] must lie in the tangent line. We may as well pick \(t=1\) to get a nonzero vector in the tangent line: \[ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}. \] This vector points in the direction that the curve is traced out as \(t\) increases, so it is a positively-oriented tangent vector. To make it a unit vector, we divide by its length: \[ \mathbf{t} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix}. \] The graph is here.

Exercise 4: Using derivatives to approximate changes

Recall from the class that the derivative of a function \(f: \mathbb{R}^n \to \mathbb{R}^m\) at a point \(\mathbf{v}_0\) can be used to approximate changes in the value of \(f\) near \(\mathbf{v}_0\). Specifically, for \(\mathbf{h}\) small, we have \[ f(\mathbf{v}_0 + \mathbf{h}) - f(\mathbf{v}_0) \approx f'(\mathbf{v}_0) \mathbf{h}. \]

For each of the six functions in Exercise 1, use the derivative to approximate the following changes in function values. Compare your approximations to the true values.

  1. \(f: \mathbb{R}^2 \to \mathbb{R}\) given by \(f(x,y) = x^2 - y^2\). Compute \(f'(x,y)\).
  2. \(g: \mathbb{R}^2 \to \mathbb{R}\) given by \(g(x,y) = 4-x^2-4y^2\). Compute \(g'(x,y)\).
  3. \(q: \mathbb{R} \to \mathbb{R}^2\) given by \(q(t) = (t + 2t^2, -t^3)\). Compute \(q'(t)\).
  4. \(r: \mathbb{R} \to \mathbb{R}^3\) given by \(r(t) = (e^t, \sin{t}, \cos{t})\). Compute \(r'(t)\).
  5. \(u:\mathbb{R}^3 \to \mathbb{R}\) given by \(u(x,y,z) = x^2 + y^2 + z^2\). Compute \(u'(x,y,z)\).
  6. \(p: \mathbb{R}^2 \to \mathbb{R}^3\) given by \(p(x,y) = (x^2, 3y, y^2)\). Compute \(p'(x,y)\).
  1. \(f(1.2,0.98) - f(1.1,1.0)\).
  2. \(g(1.17,0.98) - g(1.15,1.01)\).
  3. \(q(1.24) - q(1.2)\).
  4. \(r(5.35) - r(5.26)\).
  5. \(u(4.2,0.32,18.5) - u(4.13,0.42,18.3)\).
  6. \(p(-21.2,0.34) - p(-21.8,0.33)\).

  1. True value: \(0.2696\). Approximation: \(0.26\).

  2. True value: \(0.1924\). Approximation: \(0.1964\).

  3. True value: \(\begin{bmatrix} 0.2352 \\ -0.178624\end{bmatrix}\). Approximation: \(\begin{bmatrix} 0.232 \\ -0.1728 \end{bmatrix}\).

  4. True value: \(\begin{bmatrix}18.1268 \\ 0.0502506 \\ 0.0746285\end{bmatrix}\). Approximation: \(\begin{bmatrix} 17.3233 \\ 0.0468584 \\ 0.0768394 \end{bmatrix}\).

  5. True value: \(7.8691\). Approximation: \(7.8142\).

  6. True value: \(\begin{bmatrix}-25.8 \\ 0.03 \\ 0.0067\end{bmatrix}\). Approximation: \(\begin{bmatrix}-26.16 \\ 0.03 \\ 0.0066\end{bmatrix}\).