Question

In which direction is the curve \(x=-2 \sin t, y=2 \cos t,\) for \(0

Short answer

Answer: The curve is generated in a counterclockwise direction, moving through the southwest, northwest, northeast, and southeast quadrants.

Step-by-step solution

Find the derivative of x(t) with respect to t

Given the equation for x as \(x = -2\sin{t}\). To find \(\frac{dx}{dt}\), we take the derivative of x with respect to t. \(\frac{dx}{dt} = -2\cos{t}\)

Find the derivative of y(t) with respect to t

Given the equation for y as \(y = 2\cos{t}\). To find \(\frac{dy}{dt}\), we take the derivative of y with respect to t. \(\frac{dy}{dt} = -2\sin{t}\)

Analyze the signs of dx/dt and dy/dt

Since we need the direction of the curve for the given range of t (0<t<2π), let's determine the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) in this interval. For \(\frac{dx}{dt}\), we have \(-2\cos{t}\). The cosine function is positive in the 1st and 4th quadrants, i.e., for 0<t<π/2 and 3π/2<t<2π. Thus, dx/dt is negative for 0<t<π/2 and 3π/2<t<2π. For \(\frac{dy}{dt}\), we have \(-2\sin{t}\). The sine function is positive in the 1st and 2nd quadrants, i.e., for 0<t<π. Thus, dy/dt is negative for 0<t<π.

Determine the direction of the curve

Based on the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) in each quadrant for the given range of t (0<t<2π), we have the following: - For 0<t<π/2, both dx/dt and dy/dt are negative, so the curve goes in the southwest direction - For π/2<t<π, dx/dt is positive, and dy/dt is negative, so the curve goes in the northwest direction - For π<t<3π/2, dx/dt is positive, and dy/dt is positive, so the curve goes in the northeast direction - For 3π/2<t<2π, dx/dt is negative, and dy/dt is positive, so the curve goes in the southeast direction Therefore, within the given range of t (0<t<2π), the curve is generated in a counterclockwise direction, moving through the southwest, northwest, northeast, and finally southeast quadrants.

Key concepts

Curve Direction

Understanding the direction in which a parametric curve is traced is essential to knowing how it behaves. When dealing with parametric equations like \(x = -2 \sin t\) and \(y = 2 \cos t\), the direction can be determined by looking at the signs of their derivatives. With respect to the parameter \(t\), these derivatives tell us how the curve is moving.

For \(0 < t < 2\pi\), analyzing whether these derivatives are positive or negative in different quadrants gives insight into the curve's progression. The direction can also be visualized by tracing out small segments of the curve as \(t\) changes. By observing these segments over a complete cycle from \(t = 0\) to \(t = 2\pi\), it's clear that the curve is moving counterclockwise across the coordinate plane.

It's worth noting that the starting point at \(t = 0\) corresponds to the position \((-2, 0)\), and as \(t\) increases towards \(2\pi\), the curve loops around back to the starting position, forming a circle.

Trigonometric Derivatives

Trigonometric derivatives play a crucial role in analyzing parametric equations. Given \(x = -2\sin t\) and \(y = 2\cos t\), we want to find how these equations change with time, or more precisely, with \(t\). This requires calculating the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

• The derivative of \(x\) with respect to \(t\) is \(-2\cos t\), revealing how \(x\) changes as \(t\) changes.
• The derivative of \(y\) is \(-2\sin t\), showing how \(y\) changes with \(t\).

Together, these derivatives form the rate of change vector \((-2\cos t, -2\sin t)\). By understanding the nature of \(\cos t\) and \(\sin t\), which have periodic values, we can interpret the curvature and motion dynamics of the parametric curve.

Quadrant Analysis

Quadrant analysis is all about understanding how trigonometric functions behave within their specific quadrants and using this knowledge to determine the behavior of a parametric curve.

The unit circle, divided into four quadrants, shows the distribution of positive and negative signs for \(\sin t\) and \(\cos t\):

• In the 1st quadrant (\(0 < t < \frac{\pi}{2}\)), both \(\sin t\) and \(\cos t\) are positive.
• In the 2nd quadrant (\(\frac{\pi}{2} < t < \pi\)), \(\sin t\) is positive while \(\cos t\) is negative.
• In the 3rd quadrant (\(\pi < t < \frac{3\pi}{2}\)), both \(\sin t\) and \(\cos t\) become negative.
• In the 4th quadrant (\(\frac{3\pi}{2} < t < 2\pi\)), \(\cos t\) is positive and \(\sin t\) is negative.

By applying these principles, we can predict how the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) vary across the quadrants, influencing the curvilinear movement and allowing us to deduce the overall direction of the curve.