AOS5 Topic 8: Vector Calculus

Then \( \overrightarrow{PQ} = r(t + h) - r(t) \).

It follows that \( \frac{1}{h} (\overrightarrow{PQ}) \) is a vector parallel to \( \overrightarrow{PQ} \).

As \( h \rightarrow 0 \), the point \( Q \) approaches \( P \) along the curve.

The derivative of \( r \) with respect to \( t \) is denoted by \( \dot{r} \) and is defined by

\[ \dot{r}(t) = \lim_{{h \to 0}} \frac{r(t + h) - r(t)}{h} \]

provided that this limit exists.

The vector \( \dot{r}(t) \) points along the tangent to the curve at \( P \), in the direction of increasing \( t \).

Note: The derivative of a vector function \( r(t) \) is also denoted by \( \frac{dr}{dt} \) or \( r'(t) \).

Derivative of a vector function:

Let \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} \). If both \( x(t) \) and \( y(t) \) are differentiable, then

\( \mathbf{r}'(t) = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j} \)

Proof: By the definition, we have

\( \mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t + h) - \mathbf{r}(t)}{h} \)

\( = \lim_{h \to 0} \frac{(x(t + h)\mathbf{i} + y(t + h)\mathbf{j}) - (x(t)\mathbf{i} + y(t)\mathbf{j})}{h} \)

\( = \lim_{h \to 0} \frac{x(t + h)\mathbf{i} - x(t)\mathbf{i}}{h} + \lim_{h \to 0} \frac{y(t + h)\mathbf{j} - y(t)\mathbf{j}}{h} \)

\( = \left( \lim_{h \to 0} \frac{x(t + h) - x(t)}{h} \right) \mathbf{i} + \left( \lim_{h \to 0} \frac{y(t + h) - y(t)}{h} \right) \mathbf{j} \)

\( \mathbf{r}'(t) = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j} \)

Hence, proved.

Second Derivative of a Vector Function

The second derivative of \( \mathbf{r}(t) \) is

\[ \mathbf{r}''(t) = \frac{d^2x}{dt^2}\mathbf{i} + \frac{d^2y}{dt^2}\mathbf{j} = x''(t)\mathbf{i} + y''(t)\mathbf{j} \]

This can be extended to three-dimensional vector functions:

\[ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \] \[ \mathbf{r}'(t) = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j} + \frac{dz}{dt}\mathbf{k} \] \[ \mathbf{r}''(t) = \frac{d^2x}{dt^2}\mathbf{i} + \frac{d^2y}{dt^2}\mathbf{j} + \frac{d^2z}{dt^2}\mathbf{k} \]

We have the following results for differentiating vector functions:

Properties of the Derivative of a Vector Function:

  • \(\frac{d}{dt}(c) = 0\), where \(c\) is a constant vector.
  • \(\frac{d}{dt}(kr(t)) = k \frac{d}{dt}(r(t))\), where \(k\) is a real number.
  • \(\frac{d}{dt}(r_1(t) + r_2(t)) = \frac{d}{dt}(r_1(t)) + \frac{d}{dt}(r_2(t))\).
  • \(\frac{d}{dt}(f(t) \cdot r(t)) = f(t) \frac{d}{dt}(r(t)) + \frac{d}{dt}(f(t)) \cdot r(t)\), where \(f\) is a real-valued function.

    • Antidifferentiation:

    Consider \(\int r(t) \, dt = \int x(t) \, i + y(t) \, j + z(t) \, k \, dt\)

    \(= \left[ \int x(t) \, dt \right] i + \left[ \int y(t) \, dt \right] j + \left[ \int z(t) \, dt \right] k\)

    \(= X(t) \, i + Y(t) \, j + Z(t) \, k + c\)

    where \(\frac{dX}{dt} = x(t)\), \(\frac{dY}{dt} = y(t)\), \(\frac{dZ}{dt} = z(t)\), and \(c\) is a constant vector. Note that \(\frac{dc}{dt} = 0\).

    Example 1

    Derivatives of a Vector Function

    If \( \mathbf{r}(t) = t\mathbf{i} + (t - 1)^3 + 1\mathbf{j} \), find \( \mathbf{r}'(\alpha) \) and \( \mathbf{r}''(\alpha) \), where \( \mathbf{r}(\alpha) = \mathbf{i} + \mathbf{j} \).

    Solution:

    \( \mathbf{r}(t) = t\mathbf{i} + (t - 1)^3 + 1\mathbf{j} \)

    \( \mathbf{r}'(t) = \mathbf{i} + 3(t - 1)^2\mathbf{j} \)

    \( \mathbf{r}''(t) = 6(t - 1)\mathbf{j} \)

    We have \( \mathbf{r}(\alpha) = \alpha\mathbf{i} + (\alpha - 1)^3 + 1\mathbf{j} = \mathbf{i} + \mathbf{j} \)

    Therefore, \( \alpha = 1 \), and \( \mathbf{r}'(1) = \mathbf{i} \) and \( \mathbf{r}''(1) = \)

    Example 2

    Finding derivative of vector function nd gradient of curve

    A curve is described by the vector equation \( \mathbf{r}(t) = 2 \cos(t) \mathbf{i} + 3 \sin(t) \mathbf{j} \).

    a) Find:

    1. \( \mathbf{r}'(t) \)
    2. \( \mathbf{r}''(t) \)

    b) Find the gradient of the curve at the point \( (x, y) \), where \( x = 2 \cos(t) \) and \( y = 3 \sin(t) \).

    Solution:

    a) i) \( \mathbf{r}'(t) = -2 \sin(t) \mathbf{i} + 3 \cos(t) \mathbf{j} \)

    ii) \( \mathbf{r}''(t) = -2 \cos(t) \mathbf{i} - 3 \sin(t) \mathbf{j} \)

    b) We can find \( \frac{{dy}}{{dx}} \) using related rates:

    \( \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}, \frac{{dx}}{{dt}} = -2 \sin(t) \), \( \frac{{dy}}{{dt}} = 3 \cos(t) \)

    \( \frac{{dy}}{{dx}} = \frac{{3 \cos(t)}}{{-2 \sin(t)}} = -\frac{3}{2} \cot(t) \)

    Note that the gradient is undefined when \( \sin(t) = 0 \)

    Example 3

    Gradient of a Vector Curve

    A curve is described by the vector equation \( \mathbf{r}(t) = \sec(t) \mathbf{i} + \tan(t) \mathbf{j} \), with \( t \in [-\frac{\pi}{2}, \frac{\pi}{2}] \).

    a) Find the gradient of the curve at the point \( (x, y) \), where \( x = \sec(t) \) and \( y = \tan(t) \).

    b) Find the gradient of the curve where \( t = \frac{\pi}{4} \).

    Solution:

    a) \( x = \sec(t) = \frac{1}{\cos(t)} = (\cos(t))^{-1} \) and \( y = \tan(t) \)

    \( \frac{{dx}}{{dt}} = -(\cos(t))^{-2}(-\sin(t)) \)

    \( \frac{{dy}}{{dt}} = \sec^2(t) = \frac{{\sin(t)}}{{\cos^2(t)}} = \tan(t) \sec(t) \)

    Hence \( \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \sec^2(t) \cdot \frac{1}{{\tan(t) \sec(t)}} = \sec(t) \cot(t) = \frac{1}{{\sin(t)}} \)

    b) When \( t = \frac{\pi}{4} \), \( \frac{{dy}}{{dx}} = \frac{1}{{\sin(\frac{\pi}{4})}} = \sqrt{2} \)

    Example 4

    Given that \( \ddot{\mathbf{r}}(t) = 10\mathbf{i} - 12\mathbf{k} \), find:

    \( \dot{\mathbf{r}}(t) \) if \( \dot{\mathbf{r}}(0) = 30\mathbf{i} - 20\mathbf{j} + 10\mathbf{k} \)

    Solution

    a) \( \dot{\mathbf{r}}(t) = 10t\mathbf{i} - 12t\mathbf{k} + \mathbf{c}_1 \), where \( \mathbf{c}_1 \) is a constant vector

    \( \dot{\mathbf{r}}(0) = 30\mathbf{i} - 20\mathbf{j} + 10\mathbf{k} \)

    Thus \( \mathbf{c}_1 = 30\mathbf{i} - 20\mathbf{j} + 10\mathbf{k} \)

    and \( \dot{\mathbf{r}}(t) = 10t\mathbf{i} - 12t\mathbf{k} + 30\mathbf{i} - 20\mathbf{j} + 10\mathbf{k} \)

    \( = (10t + 30)\mathbf{i} - 20\mathbf{j} + (10 - 12t)\mathbf{k} \)

    Example 5

    antidifferentiation

    Given \( \mathbf{r}''(t) = -9.8 \mathbf{j} \) with \( \mathbf{r}(0) = 0 \) and \( \mathbf{r}'(0) = 30\mathbf{i} + 40\mathbf{j} \), find \( \mathbf{r}(t) \).

    Solution:

    \( \mathbf{r}''(t) = -9.8 \mathbf{j} \)

    \( \mathbf{r}'(t) = \int 0 \, dt \mathbf{i} + \int -9.8 \, dt \mathbf{j} \)

    \( = -9.8t \mathbf{j} + c_1 \)

    But \( \mathbf{r}'(0) = 30\mathbf{i} + 40\mathbf{j} \), giving \( c_1 = 30\mathbf{i} + 40\mathbf{j} \).

    \( \therefore \mathbf{r}'(t) = 30\mathbf{i} + (40 - 9.8t)\mathbf{j} \)

    \( \mathbf{r}(t) = \int 30 \, dt \mathbf{i} + \int (40 - 9.8t) \, dt \mathbf{j} \)

    Thus \( = 30t\mathbf{i} + (40t - 4.9t^2)\mathbf{j} + c_2 \)

    Now \( \mathbf{r}(0) = 0 \) and therefore \( c_2 = 0 \).

    Hence \( \mathbf{r}(t) = 30t\mathbf{i} + (40t - 4.9t^2)\mathbf{j} \).

    Exercise &&1&& (&&1&& Question)

    Find \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) if \( \mathbf{r}(t) = 20t\mathbf{i} + (15t - 5t^2)\mathbf{j} \).

    1
    Submit

    Exercise &&2&& (&&1&& Question)

    Find \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) if \( \mathbf{r}(t) = \cos(t)\mathbf{i} - \sin(t)\mathbf{j} + 5t\mathbf{k} \).

    2
    Submit

    Exercise &&3&& (&&1&& Question)

    Given that \( \ddot{\mathbf{r}}(t) = 10\mathbf{i} - 12\mathbf{k} \), find:

    \( \mathbf{r}(t) = (5t^2 + 30t)\mathbf{i} - 20t\mathbf{j} + (10t - 6t^2 + 2)\mathbf{k} + \mathbf{c}_2 \), where \( \mathbf{c}_2 \) is a constant vector

    3
    Submit

    Exercise &&4&& (&&1&& Question)

    Find the gradient at the point on the curve determined by the given value of \( t \) for the following:

    \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} \), \( t = \frac{\pi}{4} \)

    4
    Submit

    Exercise &&5&& (&&1&& Question)

    Find \( \mathbf{r}(t) \) for the following: \( \frac{d\mathbf{r}}{dt} = 4\mathbf{i} + 3\mathbf{j} \), where \( \mathbf{r}(0) = \mathbf{i} - \mathbf{j} \)

    5
    Submit