Math Problem Statement

Determine which is the graph of the curve with vector equation r(t)=⟨5t,−2sin(t)⟩ for −π2≤t≤π2.

Solution

To determine the graph of the curve given by the vector equation r(t)=5t,2sin(t)\mathbf{r}(t) = \langle 5t, -2\sin(t) \rangle for π2tπ2-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}, let's analyze the components and how they relate to the plot.

Step-by-Step Breakdown:

  1. x-component (horizontal): The x-component of the vector is x(t)=5tx(t) = 5t. This indicates a linear relationship between xx and tt, meaning the x-values are scaled by 5. As tt increases, xx will increase proportionally by 5 times the value of tt.

    Since tt is bounded by π2tπ2-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}, the corresponding x-values will range from: x(π2)=5×π2=5π2tox(π2)=5×π2=5π2.x\left( -\frac{\pi}{2} \right) = 5 \times -\frac{\pi}{2} = -\frac{5\pi}{2} \quad \text{to} \quad x\left( \frac{\pi}{2} \right) = 5 \times \frac{\pi}{2} = \frac{5\pi}{2}. So, the x-values span from 5π2-\frac{5\pi}{2} to 5π2\frac{5\pi}{2}.

  2. y-component (vertical): The y-component is y(t)=2sin(t)y(t) = -2\sin(t). This describes a sinusoidal behavior with an amplitude of 2, but it is reflected across the x-axis because of the negative sign. As tt varies from π2-\frac{\pi}{2} to π2\frac{\pi}{2}, the sine function ranges from 1-1 to 11, and thus the y-values will range from: y(π2)=2×sin(π2)=2×(1)=2,y\left( -\frac{\pi}{2} \right) = -2 \times \sin\left( -\frac{\pi}{2} \right) = -2 \times (-1) = 2, to y(π2)=2×sin(π2)=2×1=2.y\left( \frac{\pi}{2} \right) = -2 \times \sin\left( \frac{\pi}{2} \right) = -2 \times 1 = -2. Therefore, the y-values will range between -2 and 2.

Summary of the curve:

  • The x-component moves linearly from 5π2-\frac{5\pi}{2} to 5π2\frac{5\pi}{2}.
  • The y-component follows a sinusoidal oscillation between -2 and 2, with a reflection (negative sign) applied.

The resulting graph will be sinusoidal, stretched horizontally by a factor of 5 and vertically by a factor of 2, with the curve oscillating between 2 and -2 along the y-axis. This curve will have a wavelike pattern due to the sine function, but its horizontal extent is much larger due to the linear 5t5t component.

Would you like to proceed with graphing this function, or do you need further clarifications on the vector components?


Here are some questions to explore further:

  1. How would the curve change if the x-component were 3t3t instead of 5t5t?
  2. What effect does changing the amplitude of the sine function have on the graph?
  3. How would the curve look if the sine function were not reflected (i.e., positive sine)?
  4. What happens to the graph if the parameter range for tt is extended?
  5. How can you express this curve in parametric form for different intervals of tt?

Tip: For vector functions, the x and y components are independent, and analyzing them separately can help visualize how the curve behaves overall.

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Math Problem Analysis

Mathematical Concepts

Vector Functions
Parametric Equations
Trigonometric Functions

Formulas

x(t) = 5t
y(t) = -2sin(t)

Theorems

Sine Function Properties
Parametric Curve Analysis

Suitable Grade Level

Grades 10-12