Question

Find \(\|\vec{r}(t)\|\). $$ \vec{r}(t)=\langle 5 \cos t, 3 \sin t\rangle $$

Short answer

\(\|\vec{r}(t)\| = \sqrt{25 \cos^2 t + 9 \sin^2 t}\) is the magnitude.

Step-by-step solution

Understanding the Vector Function

We are given the vector function \(\vec{r}(t) = \langle 5 \cos t, 3 \sin t \rangle\). This function represents a parametric vector in a two-dimensional space with its components in terms of \( t \).

The Formula for Magnitude

To find the magnitude (or norm) of the vector \(\|\vec{r}(t)\|\), we use the formula for the magnitude of a vector \(\vec{v} = \langle a, b \rangle\), which is \(\|\vec{v}\| = \sqrt{a^2 + b^2}\).

Substitute the Components into the Formula

Substitute the components of the vector \(\vec{r}(t) = \langle 5 \cos t, 3 \sin t \rangle\) into the magnitude equation: \(\|\vec{r}(t)\| = \sqrt{(5 \cos t)^2 + (3 \sin t)^2}\).

Simplify Under the Square Root

Calculate \((5 \cos t)^2 = 25 \cos^2 t\) and \((3 \sin t)^2 = 9 \sin^2 t\). Add these two results: \(\sqrt{25 \cos^2 t + 9 \sin^2 t}\).

Simplify the Expression

Factor the terms inside the square root: there are no common factors, so we calculate each separately. The expression becomes \(\sqrt{25 \cos^2 t + 9 \sin^2 t}\). Since there is no identity to simplify more, the expression remains as is.

Key concepts

Vector Magnitude

The concept of vector magnitude is fundamental in calculus as it gives us the size or length of a vector. For any vector \( \vec{v} = \langle a, b \rangle \), its magnitude is calculated using the formula \( \|\vec{v}\| = \sqrt{a^2 + b^2} \). This equation is essentially a derived form of the Pythagorean theorem, applied to vectors. When you visualize a vector in two-dimensional space, think of it as the diagonal of a right triangle.

• The components \( a \) and \( b \) correspond to the lengths of the triangle's two perpendicular sides.
• The magnitude is the length of the hypotenuse of this right triangle.

The magnitude function turns the abstract concept of a vector into a concrete number that we can interpret as length. This allows for comparison between vectors, understanding the vector’s size regardless of its direction.

Parametric Equations

Parametric equations describe a method of expressing geometric figures using parameters. In this case, the parameter is the variable \( t \). Instead of describing the relationship between \( x \) and \( y \) with one equation like \( y=f(x) \), we use a pair of equations:
\( x(t) = 5 \cos t \)
\( y(t) = 3 \sin t \)
Each equation gives either the \( x \) or \( y \) coordinate as a function of \( t \).

• The parameter \( t \) can be thought of as time, showing how the position of a point changes as time progresses.
• These equations make it easier to express curves and trajectories in a plane, especially for circles and ellipses.

In our original problem, the vector \( \vec{r}(t) = \langle 5 \cos t, 3 \sin t \rangle \) represents an ellipse as \( t \) varies. Each value of \( t \) gives a different point on the ellipse.

Trigonometric Functions

Trigonometric functions like cosine and sine play a key role in parametric equations and vectors. When the components of a vector are defined using trigonometry,it often has applications in physics and engineering.

• These functions help model periodic phenomena, giving vectors a smooth and continuous path.
• In this context, \( \cos t \) and \( \sin t \) affect how the vector traces out its path in the plane, effectively mapping out an ellipse whenever \( t \) is involved in both components.

Specific to the original problem, these trigonometric expressions scale the vector components by controlling their amplitude:
\( 5 \cos t \) means the \( x \) component oscillates between \(-5\) and \(5\),
\( 3 \sin t \) means the \( y \) component oscillates between \(-3\) and \(3\).
Understanding how trigonometric functions behave in various scenarios helps us predict and interpret the movement of vectors in contexts larger than simple geometry.