Basic Mathematics · Unit 9
Multivariable Calculus
Exam-focused notes for Multivariable Calculus (Basic Mathematics, MTH104): what the TU syllabus asks and how it has actually been tested, with 10 solved past questions from this unit.
What this unit covers
- Partial derivatives
- Second order partial derivatives
- Gradient vector
- Directional derivatives
- Chain rule for multivariable functions
- Implicit differentiation in multiple variables
- Local extrema of multivariable functions
- Critical points and classification
Partial derivatives
Find the partial derivatives $f_x$, $f_y$ and $f_{xy}$ of $f(x,y) = \sqrt{x} y^3 + x^4 y$ at $(-4,1)$. [5]
- Function: $f(x,y) = \sqrt{x}\, y^3 + x^4 y$ - Evaluation point: $(-4, 1)$ --- Treat $y$ as constant. Write $\sqrt{x} = x^{1/2}$. $$fx = \frac{1}{2}x^{-1/2}\,y^3 + 4x^3 y = \frac{y^3}{2\sqrt{x}} + 4x^3 y$$ --- Treat $x$ as constant. $$fy = 3\sqrt{x}\,y^2 +...
Full solved answer →Define partial derivative and find the value of $\frac{\partial f}{\partial x}$ & $\frac{\partial f}{\partial y}$ at the point (4, -5) if $f(x,y) = x^3 + 3xy + y - 1$. [5]
- Function: $f(x, y) = x^3 + 3xy + y - 1$ - Point of evaluation: $(x, y) = (4, -5)$ - Required: $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ at $(4, -5)$ A partial derivative of a function $f(x, y)$ with respect to one variable is i...
Full solved answer →Gradient vector
(question text pending review) # Solution
Part 1: Gradient and Directional Derivative [5 marks]
Definition of Gradient: The gradient of a scalar function $f(x,y,z)$ is defined as: $$
Part 1: - Function: $f(x,y,z) = x^3 - xy^2 + z$ - Point: $P(1,0,0)$ - Direction: $\vec{v} = 2\vec{i} - \vec{j} + \vec{k}$ Part 2: - Region bounded by $y = \sqrt{x}$, line $y = 1$, and $x = 4$ - Axis of revolution: $y = 1$ All data present and readable. --- ...
Full solved answer →Question
Define Gradient vector and directional derivative. Find the direction in which $f(x,y) = \frac{x^2}{2} + \frac{y^2}{2}$ increases and decreases most rapidly at the point $(1,1)$. What is the direction of zero change in $f$ at $(1,1)$? Derivative of $f(x,y)$ at the point $(1,1)$ in the direction $v = 3i - 4j$. [5+2+3]
- Function: $f(x,y) = \dfrac{x^2}{2} + \dfrac{y^2}{2}$ - Point: $(1,1)$ - Direction vector for part 3: $\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}$ --- (a) Definitions The gradient of $f(x,y)$ is the vector of its first partial derivatives: $$\nabla f = \left\l...
Full solved answer →Chain rule for multivariable functions
What is chain rule for function $w = f(x,y)$? To use this rule, find the derivative of $w = xy$ w.r.t. $t$ along the path $x = \cos t$, $y = \sin t$. Also, find derivative of $w$ at $t = \frac{\pi}{2}$. [1+4]
- $w = f(x,y) = xy$ - $x = \cos t$, $y = \sin t$ - Evaluate $\dfrac{dw}{dt}$ at $t = \dfrac{\pi}{2}$ --- For $w = f(x, y)$ where $x = x(t)$ and $y = y(t)$ are functions of a single variable $t$, the chain rule gives the total derivative: $$\frac{dw}{dt} = \...
Full solved answer →Directional derivatives
Find the derivative of $f(x, y, z) = x^3 - xy^2 - z$ at point $P(1, 1, 0)$ in the direction of $v = 2i - 3j + 6k$. [5]
- Function: $f(x,y,z) = x^3 - xy^2 - z$ - Point: $P(1, 1, 0)$ - Direction vector: $\mathbf{v} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ The directional derivative is $D{\mathbf{u}}f = \nabla f \cdot \hat{\mathbf{u}}$, where $\hat{\mathbf{u}}$ is the unit v...
Full solved answer →Find the derivatives of the function $f(x,y) = x^3 - xy^2 + x^2y - y^3$ at the point $p_0(5,5)$ in the direction of $\vec{u} = 4\vec{i} + 3\vec{j}$. [5]
- Function: $f(x,y) = x^3 - xy^2 + x^2y - y^3$ - Point: $p0(5,5)$ - Direction: $\vec{u} = 4\vec{i} + 3\vec{j}$ $$D{\vec{u}}f = \nabla f \cdot \hat{u}$$ $$\frac{\partial f}{\partial x} = 3x^2 - y^2 + 2xy$$ $$\frac{\partial f}{\partial y} = -2xy + x^2 - 3y^2$...
Full solved answer →Implicit differentiation in multiple variables
Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ if $x^3 + y^3 + z^3 + 6xyz = 1$. [5]
Implicit relation: $$F(x,y,z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0$$ Required: $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$, where $z$ is an implicit function of $x$ and $y$. No numeric values are missing; this is a symbolic derivation....
Full solved answer →Second order partial derivatives
Find the second order derivative $\frac{\partial^2 f}{\partial x^2}, \frac{\partial^2 f}{\partial y^2}, \frac{\partial^2 f}{\partial x\partial y}, \frac{\partial^2 f}{\partial y\partial x}$ of $f(x,y) = x \cos y + ye^x$ [5]
$$f(x,y) = x\cos y + ye^x$$ Required: $\dfrac{\partial^2 f}{\partial x^2}, \dfrac{\partial^2 f}{\partial y^2}, \dfrac{\partial^2 f}{\partial x\,\partial y}, \dfrac{\partial^2 f}{\partial y\,\partial x}$ Differentiate with respect to $x$ (treat $y$ constant)...
Full solved answer →Local extrema of multivariable functions
Find the local extreme values of the function $f(x,y) = xy - x^2 - y^2 - 2x - 2y + 4$. [5]
Function: $f(x,y) = xy - x^2 - y^2 - 2x - 2y + 4$ First partial derivatives: $$fx = y - 2x - 2$$ $$fy = x - 2y - 2$$ Set equal to zero: $$y - 2x - 2 = 0 \quad (1)$$ $$x - 2y - 2 = 0 \quad (2)$$ From (1): $y = 2x + 2$ Substitute into (2): $$x - 2(2x+2) - 2 =...
Full solved answer →Make Unit 9 stick
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