Problem 39 - Precalculus Mathematics for Calculus

Question

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$\mathbf{u}=\langle 10,-1\rangle, \quad \mathbf{v}=\langle- 2,-2\rangle$$

Step-by-Step Solution

Answer

\(|\mathbf{u}| = \sqrt{101}, |\mathbf{v}| = 2\sqrt{2}, |2\mathbf{u}| = 2\sqrt{101}, \left|\frac{1}{2}\mathbf{v}\right| = \sqrt{2}, |\mathbf{u}+\mathbf{v}| = \sqrt{73}, |\mathbf{u}-\mathbf{v}| = \sqrt{145}, |\mathbf{u}|-|\mathbf{v}| = \sqrt{101} - 2\sqrt{2}\).

Step 1: Find the magnitude of \( \mathbf{u} \)

The magnitude of a vector \( \mathbf{u} = \langle x, y \rangle \) is given by \( |\mathbf{u}| = \sqrt{x^2 + y^2} \). Substitute \( x = 10 \) and \( y = -1 \) to find the magnitude:\[ |\mathbf{u}| = \sqrt{10^2 + (-1)^2} = \sqrt{100 + 1} = \sqrt{101} \approx 10.05 \]

Step 2: Find the magnitude of \( \mathbf{v} \)

The magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \) is given by \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). Substitute \( x = -2 \) and \( y = -2 \) to find the magnitude:\[ |\mathbf{v}| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \]

Step 3: Find the magnitude of \( 2\mathbf{u} \)

Multiplying a vector by a scalar multiplies its magnitude by the absolute value of the scalar. Therefore, the magnitude of \( 2\mathbf{u} \) is:\[ |2\mathbf{u}| = 2|\mathbf{u}| = 2\sqrt{101} \approx 2 \times 10.05 = 20.1 \]

Step 4: Find the magnitude of \( \frac{1}{2}\mathbf{v} \)

Similarly, the magnitude of \( \frac{1}{2}\mathbf{v} \) is half the magnitude of \( \mathbf{v} \):\[ \left| \frac{1}{2}\mathbf{v} \right| = \frac{1}{2}|\mathbf{v}| = \frac{1}{2}\times 2\sqrt{2} = \sqrt{2} \approx 1.41 \]

Step 5: Find the magnitude of \( \mathbf{u} + \mathbf{v} \)

Find \( \mathbf{u} + \mathbf{v} = \langle 10, -1 \rangle + \langle -2, -2 \rangle = \langle 8, -3 \rangle \). Then compute the magnitude:\[ |\mathbf{u} + \mathbf{v}| = \sqrt{8^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73} \approx 8.54 \]

Step 6: Find the magnitude of \( \mathbf{u} - \mathbf{v} \)

First, find \( \mathbf{u} - \mathbf{v} = \langle 10, -1 \rangle - \langle -2, -2 \rangle = \langle 12, 1 \rangle \). Then calculate the magnitude:\[ |\mathbf{u} - \mathbf{v}| = \sqrt{12^2 + 1^2} = \sqrt{144 + 1} = \sqrt{145} \approx 12.04 \]

Step 7: Calculate \( |\mathbf{u}| - |\mathbf{v}| \)

Subtract the magnitude of \( \mathbf{v} \) from the magnitude of \( \mathbf{u} \):\[ |\mathbf{u}| - |\mathbf{v}| = \sqrt{101} - 2\sqrt{2} \approx 10.05 - 2.83 = 7.22 \]

Key Concepts

Vector Addition

Vector addition is a fundamental concept in vector mathematics. When you add two vectors, you essentially bring together their respective components. For instance, if you're given vectors \( \mathbf{u} = \langle x_1, y_1 \rangle \) and \( \mathbf{v} = \langle x_2, y_2 \rangle \), the sum is \( \mathbf{u} + \mathbf{v} = \langle x_1 + x_2, y_1 + y_2 \rangle \). This means you add the x-components to get the new x-component, and the y-components for the new y-component.

• Helps in determining the combined effect of two vector forces.
• Results in another vector.

By understanding vector addition, you extend your ability to solve physics and engineering problems, where multiple forces or velocities combine. In your problem, the calculation \( \mathbf{u} + \mathbf{v} = \langle 10, -1 \rangle + \langle -2, -2 \rangle = \langle 8, -3 \rangle \) simplifies the vectors into a single, resultant vector.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a real number, called a scalar. Given a vector \( \mathbf{u} = \langle x, y \rangle \) and a scalar \( c \), the resultant vector is \( c \mathbf{u} = \langle cx, cy \rangle \).
This operation:

• Stretches or shrinks the vector.
• Keeps the direction the same if the scalar is positive.
• Reverses the direction if the scalar is negative.

By controlling the size of the vector without altering its direction, scalar multiplication is crucial in vector calculus and physics. In the provided problem, referring to \( 2\mathbf{u} \), it signifies doubling every component of \( \mathbf{u} \), making \( 2\mathbf{u} = \langle 20, -2 \rangle \), subsequently doubling the magnitude as well.

Vector Subtraction

Vector subtraction involves determining the difference between two vectors. If vectors \( \mathbf{u} = \langle x_1, y_1 \rangle \) and \( \mathbf{v} = \langle x_2, y_2 \rangle \), then their subtraction is \( \mathbf{u} - \mathbf{v} = \langle x_1 - x_2, y_1 - y_2 \rangle \).

• Visualized as connecting the tip of \( \mathbf{v} \) to the tip of \( \mathbf{u} \).
• Helps in calculating relative displacement or change in position.

The calculation in the provided problem shows \( \mathbf{u} - \mathbf{v} = \langle 10, -1 \rangle - \langle -2, -2 \rangle = \langle 12, 1 \rangle \). This operation is essential for determining changes in vector quantities, such as shifting positions or correcting courses.

Norm of Vectors

The norm of a vector, often referred to as vector magnitude, is a measure of the vector's length. For a vector \( \mathbf{u} = \langle x, y \rangle \), its norm is calculated using the formula \( |\mathbf{u}| = \sqrt{x^2 + y^2} \). The norm gives:

• A scalar representing how long the vector is.
• Helps in understanding the size or extent of vector quantities such as force or velocity.

When applied in physics and engineering, it provides the actual measure of a vector's effect, such as the total force exerted. In the given solution, calculating the norms for vectors \( \mathbf{u} \) and \( \mathbf{v} \) are fundamental steps in understanding their relative magnitudes, with outcomes of \( \sqrt{101} \) and \( 2\sqrt{2} \) respectively.