91Ó°ÊÓ

In the context of parametric equations, a horizontal tangent line indicates that the slope of the curve is zero at a particular point. It's like the curve is momentarily moving perfectly flat, parallel to the x-axis.
To find when a horizontal tangent occurs, you need to determine when the derivative of the y-component with respect to the x-component, denoted as \( \frac{dy}{dx} \), equals zero. For parametric equations where \( x = 2 \sin t \) and \( y = 4 \cos t \), this involves setting \( \frac{dy}{dt} = 0 \) while ensuring that \( \frac{dx}{dt} eq 0 \).

• Start by finding the derivatives: \( \frac{dy}{dt} = -4 \sin t \) and \( \frac{dx}{dt} = 2 \cos t \).
• For horizontal tangents, set \( -4 \sin t = 0 \), giving \( \sin t = 0 \), which means \( t = 0, \pi, 2\pi \).
• Verify \( \cos t \) at these values: it should not be zero (here it's 1, -1, and 1, respectively).

These conditions ensure that at \( t = 0, \pi, 2\pi \), you have horizontal tangent lines because the curve is flat at these times.

Vertical tangent lines occur when a curve is positioned vertically, which means at that position, the change in the x-component with respect to time, \( \frac{dx}{dt} \), is zero, while the y-component is changing, making the tangent line parallel to the y-axis.
To identify where vertical tangents occur for the parametric curve described by \( x = 2 \sin t \) and \( y = 4 \cos t \), we look for situations where \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} eq 0 \).

• Given \( \frac{dx}{dt} = 2 \cos t \), solve for \( \cos t = 0 \), resulting in \( t = \frac{\pi}{2}, \frac{3\pi}{2} \).
• Then, check \( \sin t = 1, -1 \) (non-zero) for these values using \( \frac{dy}{dt} = -4 \sin t \).

By meeting these conditions, you can conclude that the parametric curve has vertical tangents at \( t = \frac{\pi}{2} \) and \( t = \frac{3\pi}{2} \), where the x-component stays constant while the y-component changes.

Derivatives in mathematics measure how a function changes as its input changes. They're like the speedometer of a function, telling us how fast something is moving.
In parametric equations where both x and y depend on a third variable, usually time \( t \), derivatives give us the rate of change of x, \( \frac{dx}{dt} \), and y, \( \frac{dy}{dt} \), with respect to time. Together, these rates help us determine things like the slope of the curve.

• \( \frac{dx}{dt} \) tells how fast the x-component is changing. For \( x = 2 \sin t \), \( \frac{dx}{dt} = 2 \cos t \).
• \( \frac{dy}{dt} \) shows the rate of change for the y-component. For \( y = 4 \cos t \), \( \frac{dy}{dt} = -4 \sin t \).

These derivatives are crucial in finding where the curve has horizontal or vertical tangents. They enable you to compute \( \frac{dy}{dx} \), ensuring you understand the curve's behavior at any point \( t \). By examining these rates, you can figure out when the tangent to the curve becomes purely horizontal or vertical.