Lecture Notes Summary on Multivariable Calculus

MATH1011 Multivariable Calculus – Sem. 1, 2026 Lecture Notes Week 4

3.10 Directional Derivatives and Gradient Vectors (Cont.)

• 3.10.1 Theorem: If $f$ is a differentiable function at $c$, then for every non-zero vector $v$, the directional derivative $D_vf(c)$ exists and is given by:

  • $D_vf(c) = <br>abla f(c) \bullet u$, where $u = rac{v}{|v|}$.

Chain Rule in Vector Form

• 3.10.2 Chain Rule:

  • Let $f(x), x = (x<em>1, ext{…}, x</em>n) ext{ in } \mathbb{R}^n$ be a real-valued function defined and differentiable in a subset $D$ of $\mathbb{R}^n$, and let vector-valued function $r(t) = (r<em>1(t), ext{…}, r</em>n(t))$ be differentiable and take values in $D$.

  • The Chain Rule states:

  • $\frac{d}{dt}f(r(t)) = <br>abla f(r(t)) \bullet r ' (t)$.

Proof of Theorem 3.10.1 (Not Examinable)

• Consider the vector-valued function:

  • $r(h) = c + hu$ defined for $h ext{ in } \mathbb{R}$ with sufficiently small $|h|$.

  • Notably, $r(0) = c$ and $r ' (h) = u$.

  • Consequently, we derive:

  • $D<em>vf(c) = \lim</em>{h \to 0} \frac{f(c + hu) - f(c)}{h} = \frac{d}{dh} f(r(h))|_{h=0} = <br>abla f(c) \bullet r ' (0) = <br>abla f(c) \bullet u$, thus proving the theorem.

Example Set 1 – Week 4

1. Find $\nabla f$ for the function $f(x, y) = x^2 + y - 2xy - 1$.

2. For the function $f(x, y) = 10 + x + y + xy$, find the rate of change of $f$ at the point $P = (1, 2)$ in the direction of the vector $v = (3, 4)$.

3. Suppose the temperature at the point $(x, y, z)$ in solid $S$ is given by $T(x, y, z) = 10 + xy + xz + yz$. Find the rate of change of the temperature at point $P = (1, 2, 3)$ in the direction of the vector $v = (1, 2, -2)$.

Maximum and Minimum Rates of Change

• To determine in which direction $f$ has maximum or minimum rates of change, recall the directional derivative:

  • $D_vf(c) = <br>abla f(c) \bullet u$ with $u$ being the unit vector in the direction of $v$.

• The dot product can also be expressed as:

  • $<br>abla f(c) \bullet u = |<br>abla f(c)||u| ext{cos } \theta = |<br>abla f(c)| ext{cos } \theta$, where $\theta$ is the angle between $\nabla f(c)$ and $u$.

Theorems on Rates of Change

• 3.10.3 Theorem: Let $f(x)$ for $x ext{ in } D$ be a differentiable function of two or more variables and let $c$ be an interior point of $D$ such that $\nabla f(c) \neq 0$.

  • Then:

  • $ext{max}<em>{|u|=1} D</em>u f(c) = |<br>abla f(c)|$ - maximum when $u = rac{\nabla f(c)}{|<br>abla f(c)|}$.

  • $ext{min}<em>{|u|=1} D</em>u f(c) = - |<br>abla f(c)|$ - minimum when $u = - rac{\nabla f(c)}{|<br>abla f(c)|}$.

Gradient Vector Implications

• The gradient vector at a point indicates the direction of greatest increase for the function.

• The function decreases most rapidly when the angle between $v$ and $\nabla f(c)$ is $\theta = \pi$ (i.e., when $v$ is in direction of $-\nabla f(c)$).

Examples and Applications

• 3.10.4 Example: The temperature at each point $(x, y)$ of a metal plate is given by:

  • $T(x, y) = e^x ext{cos } y + e^y ext{cos } x$.

  • Find the direction of rapid temperature increase at $P = (0,0)$ and the corresponding rate of increase.

• Calculation shows:

  • $\nabla T = (e^x ext{cos } y - e^y ext{sin } x, -e^x ext{sin } y + e^y ext{cos } x)$, yielding:

  • $\nabla T(0, 0) = (1, 1)$, representing the direction of fastest increase at $P = (0, 0)$.

  • The rate of increase in direction of $\nabla T(0, 0)$ is $|\nabla T(0, 0)| = \sqrt{2}$.

Example Set 2 – Week 4

1. Suppose the temperature at any point $(x, y, z)$ in a rectangular box is given by:

   • $T(x, y, z) = xyz(1 - x)(2 - y)(3 - z)$.

   • Determine the direction of flight for a mosquito situated at $P = (\frac{1}{2}, 1, 1)$ to cool off rapidly.

Relationship Between Gradient Vector and Level Curves

• Assume $f(x, y)$ is defined over a differentiable subset $D$ of $\mathbb{R}^2$. A set of the form:

  • $C = {(x, y) \in D : f(x, y) = k}$, where $k$ is a constant, is termed a level curve of $f$.

• 3.10.5 Theorem: If $f : D \to \mathbb{R}$ is differentiable at $(a, b)$ and $\nabla f(a, b) \neq 0$, then:

  • $\nabla f(a, b)$ is perpendicular to the tangent line to the level curve of $f$ at $(a, b)$.

Visual Representation

• Picture demonstrating level curves and gradient vectors:

  • Suppose a function $z = f(x,y)$ at point $(a,b)$ having a gradient vector $\nabla f(a,b)$.

  • The gradient $\nabla f(a,b)$ is shown as a normal vector to the level curve, which is perpendicular to the tangent line of the curve at that point.

Extension to Functions of Three Variables

• Let $f(x) : D \to \mathbb{R}$ with $D \subset \mathbb{R}^n$ where n ≥ 3. The level surface defined by:

  • $S : f(x) = k$,

  • $\text{Theorem 3.10.6:}$

  • The gradient $\nabla f(c)$ acts as a normal vector at point $c$, creating a tangent plane orthogonal to $\nabla f(c)$.

• A normal vector $n$ is determined via the cross product from two surrounding vectors:

  • $n = (−fx(a,b), −fy(a,b), 1)$, linking gradient vectors back to surfaces drawn through implicit functions.

Example Set 3 – Week 4

1. Find an equation for the tangent plane to the surface:

   • $S_1 : x^2 - 4y^2 + z^2 = 16$ at point $P = (2, 1, 4)$.

4. Maxima and Minima

4.1 Functions of a Single Variable

• Increasing and Decreasing Functions:

  • A function $f(x)$ is increasing (strictly increasing) on set $A$ if $f(x<em>1) ≤ f(x</em>2)$ (strictly f(x1) < f(x2)) for all $x<em>1, x</em>2 ext{ in } A$ where x1 < x2.

  • A function is decreasing (strictly decreasing) on set $A$ if $f(x<em>1) ≥ f(x</em>2)$ (strictly f(x1) > f(x2)) under the same conditions.

Important Example

• Consider function defined by:

  • $f(x) = [x]$,
    where $[x]$ is the integer part of $x$.

  • Such a function is increasing but not strictly increasing; students are encouraged to visualize this with a sketch.

Sufficient Conditions for Increasing/Decreasing Functions

• 4.1.1 Theorem: Let $f(x)$ be defined and differentiable over an interval $I$.

  • If $f'(x) ≥ 0$ (or f' (x) > 0), then $f$ is increasing (or strictly increasing).

  • Similarly, if $f' (x) ≤ 0$ (or f' (x) < 0), then $f$ is decreasing (or strictly decreasing).

• Remarks:

  • If $f' (x) ≥ 0$ with $f' (x) = 0$ only for finitely many x’s, then $f$ is strictly increasing.

  • The same statements hold for continuity on $I$ combined with differentiability in its interior.

4.1.2 Theorem (Mean Value Theorem):

• For the continuous function $f(x)$ on $[a, b]$, differentiable in $(a, b)$, there exists $c ext{ in } (a, b)$ such that:

  • $\frac{f(b)-f(a)}{b-a} = f'(c)$.

• Consequence implies if f'(x) > 0 on interval $I$ and x1 < x2, via MVT,

• Conclusion: f(x2) > f(x1).

4.1.3 Local Maxima and Minima

• A function $f: A \to \mathbb{R}$ has a local maximum at $c$ if:

  • $f(x) ≤ f(c)$ (or a local minimum if $f(x) ≥ f(c)$) for nearby points of $c$.

• 4.1.4 Theorem: For a differentiable function $f$ with a local maximum/minimum at interior point $c$, $f'(c) = 0$.

Visual Representation of Maxima/Minima

• If $y=f(x)$ with a local max at $c$, it indicates above zero to the left and below zero moving right; similar for minimum using reverse conditions.

4.1.5 Critical Points and Inflection Points

• Critical Point: The interior values of $f$ where $f'(c)=0$ or is undefined classify as critical points. Examples include:

  • $f(x) = x^{3/2}$ has a critical point $c = 0$; however, it's not classified as critical since it is not interior to the domain.

  • Conversely, $f(x) = |x|$ at $c=0$ is a critical point and has a local minimum.

4.1.6 Second Derivative Test

1. If f''(c) < 0, then $f(c)$ is a local maximum.

2. If f''(c) > 0, then $f(c)$ is a local minimum.

3. If $f''(c) = 0$, the test fails; either max, min or neither can hold.

4.1.7 Concavity

• If f''(x) < 0 (concave down), implies $f'$ is decreasing. Conversely, for f''(x) > 0 (concave up), $f'$ will be increasing. Inflection points occur when the concavity changes sign.

Example Analysis

• Example with $f(x)=x^3$ illustrates the behavior at point $(0,0)$ being a critical point with inflection: concave down to the left and concave up to the right.