Solutions to Calculus and Linear Algebra Comprehensive Examination

February 3, 2006

Calculus Problems and Solutions

Problem 1

Limit Problem: Calculate \(\lim_{x \to 0} \frac{\sin(3x)}{x}\)

Solution:

We can solve this limit using the standard limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\).

First, rewrite the expression: \[\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{3 \cdot \sin(3x)}{3x}\]

Let \(u = 3x\). Then as \(x \to 0\), \(u \to 0\): \[\lim_{u \to 0} \frac{3 \cdot \sin(u)}{u} = 3 \cdot \lim_{u \to 0} \frac{\sin(u)}{u}\]

Using the standard limit: \[3 \cdot 1 = 3\]

Therefore, \(\lim_{x \to 0} \frac{\sin(3x)}{x} = 3\)

Problem 2

Derivative Problem: Find the derivative of \(f(x) = \ln(\sin^2 x)\)

Solution:

We'll use the chain rule to find the derivative.

Let \(g(x) = \sin^2 x = (\sin x)^2\) and \(h(g) = \ln(g)\)

Using the chain rule: \(f'(x) = h'(g(x)) \cdot g'(x)\)

First, find \(h'(g)\): \[h'(g) = \frac{1}{g} = \frac{1}{\sin^2 x}\]

Next, find \(g'(x)\): \[g'(x) = 2(\sin x)(\cos x) = 2\sin x \cos x = \sin(2x)\]

Now, putting it together: \[f'(x) = h'(g(x)) \cdot g'(x) = \frac{1}{\sin^2 x} \cdot \sin(2x)\]

Simplifying: \[f'(x) = \frac{\sin(2x)}{\sin^2 x} = \frac{2\sin x \cos x}{\sin^2 x} = \frac{2\cos x}{\sin x} = 2\cot x\]

Therefore, \(f'(x) = 2\cot x\)

Problem 3

Integral Problem: Evaluate \(\int x^2 \ln(x) \, dx\)

Solution:

We'll use integration by parts with the formula \(\int u \, dv = uv - \int v \, du\)

Let \(u = \ln(x)\) and \(dv = x^2 \, dx\)

Then, \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^3}{3}\)

Applying integration by parts: \[\int x^2 \ln(x) \, dx = \frac{x^3}{3}\ln(x) - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx\]

Simplifying: \[\frac{x^3}{3}\ln(x) - \frac{1}{3}\int x^2 \, dx = \frac{x^3}{3}\ln(x) - \frac{1}{3} \cdot \frac{x^3}{3}\]

Adding the constant of integration: \[\frac{x^3}{3}\ln(x) - \frac{x^3}{9} + C\]

Therefore, \(\int x^2 \ln(x) \, dx = \frac{x^3}{3}\ln(x) - \frac{x^3}{9} + C\)

Problem 4

Multivariable Calculus Problem: Find the gradient of the function \(f(x,y,z) = x^2 y + yz^2 + xyz\)

Solution:

The gradient of a function \(f(x,y,z)\) is given by: \[\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)\]

Calculating each partial derivative:

\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 y + yz^2 + xyz) = 2xy + yz\]

\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 y + yz^2 + xyz) = x^2 + z^2 + xz\]

\[\frac{\partial f}{\partial z} = \frac{\partial}{\partial z}(x^2 y + yz^2 + xyz) = 2yz + xy\]

Therefore, \(\nabla f = (2xy + yz, x^2 + z^2 + xz, 2yz + xy)\)

Linear Algebra Problems and Solutions

Problem 1

Vector Space Problem: Determine whether the set \(S = \{(1,0,2), (0,1,1), (1,1,3)\}\) is linearly independent.

Solution:

To determine if a set of vectors is linearly independent, we need to check if the only solution to the equation \(c_1v_1 + c_2v_2 + c_3v_3 = 0\) is \(c_1 = c_2 = c_3 = 0\).

Setting up the equation: \[c_1(1,0,2) + c_2(0,1,1) + c_3(1,1,3) = (0,0,0)\]

This gives us the system of equations: \[\begin{cases} c_1 + c_3 = 0 \\ c_2 + c_3 = 0 \\ 2c_1 + c_2 + 3c_3 = 0 \end{cases}\]

From the first equation: \(c_1 = -c_3\)

From the second equation: \(c_2 = -c_3\)

Substituting these into the third equation: \[2(-c_3) + (-c_3) + 3c_3 = -2c_3 - c_3 + 3c_3 = 0\]

This equation is satisfied for all values of \(c_3\), meaning there are non-trivial solutions to the system.

Therefore, the set \(S\) is linearly dependent.

Problem 2

Matrix Problem: Find the inverse of the matrix \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\)

Solution:

For a 22 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the inverse is given by: \[A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\]

For our matrix \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\): \[a = 1, b = 2, c = 3, d = 4\]

Calculating the determinant: \[det(A) = ad - bc = (1)(4) - (2)(3) = 4 - 6 = -2\]

Since the determinant is non-zero, the inverse exists and is: \[A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}\]

Therefore, \(A^{-1} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}\)

Problem 3

Eigenvalue Problem: Find the eigenvalues and eigenvectors of the matrix \(A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}\)

Solution:

To find the eigenvalues, we need to solve the characteristic equation: \[det(A - \lambda I) = 0\]

Setting up the determinant: \[\begin{vmatrix} 3-\lambda & 1 \\ 1 & 3-\lambda \end{vmatrix} = 0\]

Computing the determinant: \[(3-\lambda)(3-\lambda) - 1 = 0\] \[9 - 6\lambda + \lambda^2 - 1 = 0\] \[\lambda^2 - 6\lambda + 8 = 0\]

Solving the quadratic equation: \[\lambda = \frac{6 \pm \sqrt{36-32}}{2} = \frac{6 \pm 2}{2}\] \[\lambda_1 = 4, \lambda_2 = 2\]

For \(\lambda_1 = 4\), we solve: \[(A-4I)v = 0\] \[\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

This gives us the equations \(-x + y = 0\) and \(x - y = 0\), which are equivalent to \(x = y\). Choosing \(x = 1\), we get the eigenvector \(v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

For \(\lambda_2 = 2\), we solve: \[(A-2I)v = 0\] \[\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

This gives us the equation \(x + y = 0\), which means \(x = -y\). Choosing \(x = 1\), we get the eigenvector \(v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).

Therefore, the eigenvalues are \(\lambda_1 = 4\) with eigenvector \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\lambda_2 = 2\) with eigenvector \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\).

Problem 4

Linear Transformation Problem: Let \(T: \mathbb{R}^3 \to \mathbb{R}^2\) be defined by \(T(x_1, x_2, x_3) = (2x_1 - x_2 + 3x_3, x_1 + 4x_2 - x_3)\). Find the standard matrix representation of \(T\).

Solution:

The standard matrix representation of a linear transformation \(T: \mathbb{R}^n \to \mathbb{R}^m\) is the \(m \times n\) matrix whose columns are the images of the standard basis vectors of \(\mathbb{R}^n\) under \(T\).

For \(\mathbb{R}^3\), the standard basis vectors are: \[e_1 = (1,0,0), \quad e_2 = (0,1,0), \quad e_3 = (0,0,1)\]

Finding the images of these basis vectors under \(T\):

\[T(e_1) = T(1,0,0) = (2\cdot1 - 0 + 3\cdot0, 1 + 4\cdot0 - 0) = (2,1)\]

\[T(e_2) = T(0,1,0) = (2\cdot0 - 1 + 3\cdot0, 0 + 4\cdot1 - 0) = (-1,4)\]

\[T(e_3) = T(0,0,1) = (2\cdot0 - 0 + 3\cdot1, 0 + 4\cdot0 - 1) = (3,-1)\]

Forming the matrix with these as columns: \[A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & 4 & -1 \end{pmatrix}\]

Therefore, the standard matrix representation of \(T\) is \(\begin{pmatrix} 2 & -1 & 3 \\ 1 & 4 & -1 \end{pmatrix}\).

```