Mathematics I · Unit 10 · 3 hrs
Partial Derivatives and Multiple Integrals
Exam-focused notes for Partial Derivatives and Multiple Integrals (Mathematics I, MTH117): what the TU syllabus asks and how it has actually been tested, with 14 solved past questions from this unit.
What this unit covers
- Limit and continuity
- Partial derivatives
- Tangent planes
- Maximum and minimum values
- Multiple integrals
Limit and continuity
If $f(x,y) = \frac{xy}{x^2 + y^2}$, does $\lim_{(x, y) \to (0, 0)} f(x, y)$ exist? Justify. [5]
Given data: - Function: $f(x,y) = \dfrac{xy}{x^2+y^2}$ - Point of interest: $(x,y) \to (0,0)$ - Task: determine whether the limit exists. All data present. To test a two-variable limit, approach $(0,0)$ along different paths. If limits differ, the limit doe...
Full solved answer →Partial derivatives
If $f(x,y) = 2x^3 + x^2y^2 - y^4$, find $f_x(1,-2)$, $f_y(1,-1)$ and $f_{yx}(1,-1)$. [5]
$$f(x, y) = 2x^3 + x^2y^2 - y^4$$ Required: - $fx(1, -2)$ - $fy(1, -1)$ - $f{yx}(1, -1)$ --- Differentiate with respect to $x$ (treat $y$ constant): $$fx = 6x^2 + 2xy^2 - 0 = 6x^2 + 2xy^2$$ Evaluate at $(1, -2)$: $$fx(1,-2) = 6(1)^2 + 2(1)(-2)^2 = 6 + 2(4) ...
Full solved answer →If $f(x,y) = x^3 + x^2y^3 - 2y^2$, find $f_x(2,1)$ and $f_y(2,1)$. [5]
$$f(x, y) = x^3 + x^2y^3 - 2y^2$$ Required: $fx(2,1)$ and $fy(2,1)$. --- Differentiate with respect to $x$, treating $y$ as constant: - $\frac{\partial}{\partial x}(x^3) = 3x^2$ - $\frac{\partial}{\partial x}(x^2 y^3) = 2x y^3$ - $\frac{\partial}{\partial x...
Full solved answer →Find the partial derivative $f_{xx}$ and $f_{yy}$ of $f(x,y) = x^2 + x^3y^2 - y^2 + xy$ at $(1,2)$. [5]
- Function: $f(x,y) = x^2 + x^3y^2 - y^2 + xy$ - Evaluation point: $(x, y) = (1, 2)$ - Required: $f{xx}$ and $f{yy}$ at $(1,2)$ All data is present. --- Treat $y$ as constant: $$fx = 2x + 3x^2y^2 + y$$ $$f{xx} = \frac{\partial}{\partial x}\left(2x + 3x^2y^2...
Full solved answer →Find the partial derivatives of $f(x, y) = x^2 + 2x^3y^2 - 3y^2 + x + y$ at (1,2). [5]
Function: $f(x, y) = x^2 + 2x^3y^2 - 3y^2 + x + y$ Point of evaluation: $(x, y) = (1, 2)$ Required: $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ at $(1, 2)$. --- Treat $y$ as constant: $$\frac{\partial f}{\partial x} = 2x + 2(3x^2)y...
Full solved answer →Find the partial derivative of $f(x, y) = x^3 + 2x^3y^3 - 3y^2 + x + y$ at (2,1). [5]
Function: $$f(x, y) = x^3 + 2x^3y^3 - 3y^2 + x + y$$ Point of evaluation: $(x, y) = (2, 1)$ Required: partial derivatives $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ evaluated at $(2,1)$. --- $$\frac{\partial f}{\partial x} = \frac...
Full solved answer →Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ if z is defined as a function of x and y by the equation $x^3+y^3+z^3+6xyz=1$. [5]
Equation defining $z$ implicitly as a function of $x$ and $y$: $$x^3 + y^3 + z^3 + 6xyz = 1$$ Required: $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$. Define $$F(x,y,z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0$$ By the implicit function theo...
Full solved answer →Multiple integrals
Solve $\int_0^3 \int_1^2 x^2y , dx , dy$ [5]
- Integrand: $f(x,y) = x^2 y$ - Inner integral variable: $x$, limits $[1, 2]$ - Outer integral variable: $y$, limits $[0, 3]$ Treat $y$ as constant: $$\int1^2 x^2 y \, dx = y \left[ \frac{x^3}{3} \right]1^2 = y\left(\frac{8}{3} - \frac{1}{3}\right) = \frac{...
Full solved answer →Exam Question
Part 1: Evaluate $$\int_{3}^{2} \int_{0}^{\frac{\pi}{2}} \left( y + y^{2} \cos x \right) , dx , dy$$
Part 2: Find the Maclaurin's series for $\cos x$ and prove that it represents $\cos x$ for all $x$. [5+5]
Part 1: Double integral $$\int{3}^{2}\int{0}^{\pi/2}(y + y^2\cos x)\,dx\,dy$$ - Outer limits (for $y$): from $3$ to $2$ - Inner limits (for $x$): from $0$ to $\pi/2$ Part 2: Find Maclaurin series for $\cos x$ and prove it represents $\cos x$ for all $x$. --...
Full solved answer →Question
If $f(x,y) = \frac{y}{x}$, show that $\lim_{(x,y)\to(0,0)} \frac{f(x,y)}{x}$ does not exist, justify.
Calculate $\iint_R f(x,y) dA$, for $f(x,y) = 100 - 6x^2y$, and $R: 0 \leq x \leq 2, -1 \leq y \leq 1$. [5+5]
- $f(x,y) = \dfrac{y}{x}$ - Required: examine $\displaystyle\lim{(x,y)\to(0,0)} \frac{f(x,y)}{x}$ $$\frac{f(x,y)}{x} = \frac{y/x}{x} = \frac{y}{x^2}$$ For a limit to exist, it must yield the same finite value along every path to $(0,0)$. Path 1: $y = 0$ (x-...
Full solved answer →Maximum and minimum values
Find the local maximum and minimum values, saddle points of $f(x,y) = x^4 + y^4 - 4xy + 1$. [5]
Function: $f(x,y) = x^4 + y^4 - 4xy + 1$ $$ fx = 4x^3 - 4y = 0 \Rightarrow y = x^3 \tag{1} $$ $$ fy = 4y^3 - 4x = 0 \Rightarrow x = y^3 \tag{2} $$ Substitute (1) into (2): $$x = (x^3)^3 = x^9 \Rightarrow x^9 - x = 0 \Rightarrow x(x^8 - 1) = 0$$ So $x = 0$ o...
Full solved answer →Find the local maximum and minimum values, saddle points of $f(x,y) = x^4 + y^4 - 4xy + 1$. [5]
Function: $f(x,y) = x^4 + y^4 - 4xy + 1$ $$ fx = 4x^3 - 4y = 0 \;\Rightarrow\; y = x^3 \tag{1} $$ $$ fy = 4y^3 - 4x = 0 \;\Rightarrow\; x = y^3 \tag{2} $$ Substitute (1) into (2): $$x = (x^3)^3 = x^9 \;\Rightarrow\; x^9 - x = 0 \;\Rightarrow\; x(x^8 - 1) = ...
Full solved answer →Find the extreme values of the function $f(x,y) = x^2 + 2y^2$ on the circle $x^2 + y^2 = 1$. [5]
- Objective function: $f(x,y) = x^2 + 2y^2$ - Constraint: $g(x,y) = x^2 + y^2 - 1 = 0$ Set $\nabla f = \lambda \nabla g$. $$fx = 2x,\quad fy = 4y,\qquad gx = 2x,\quad gy = 2y$$ System of equations: $$ 2x = \lambda(2x) \;\Rightarrow\; 2x(1-\lambda) = 0 \tag{...
Full solved answer →Find the extreme values of $f(x,y) = y^2 - x^2$. [5]
Function: $f(x,y) = y^2 - x^2$ No other numeric data required; this is a standard extremum problem. $$fx = -2x, \qquad fy = 2y$$ Set to zero: $$-2x = 0 \Rightarrow x = 0, \qquad 2y = 0 \Rightarrow y = 0$$ Critical point: $(0, 0)$. $$f{xx} = -2, \qquad f{yy}...
Full solved answer →Make Unit 10 stick
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