91Ó°ÊÓ

Parametric equations describe a set of related quantities as functions of an independent parameter, often denoted as 't'. In this exercise, we use the vector function \( r(t) = e^{-2t} i + \cos t \mathbf{j} + 3 \sin t \mathbf{k} \) to represent a 3D curve. The three components (\( e^{-2t}, \cos t, 3\sin t \)) are functions of \( t \) and define the x, y, and z coordinates of the curve.

• The parameter \( t \) indicates the position on the curve. By varying \( t \), you trace the entire path of the curve.
• Parametric forms have utility in describing curves that aren't functions in a traditional x-y sense.

Using parametric equations allows us to precisely control the path of the curve and is fundamental in calculus to analyze paths in vector spaces.

Vector calculus is a branch of mathematics that handles differentiation and integration of vector fields. It plays a major role in understanding vector behaviors and interactions, especially in physics and engineering. The vector function \( r(t) \) in this problem operates within a framework of vector calculus, where intersections, tangents, and derivatives are considered.

• Vectors provide direction and magnitude, and help visualize the path described by parameterizations like \( r(t) \).
• Differentiating a vector function, as shown in this exercise, involves finding the derivative of each coordinate component.

Through vector calculus, we can compute the derivative of the vector function and hence find the tangent line to a curve at a given point, as done in the steps provided.

Derivatives are a cornerstone concept in calculus, highlighting how something changes as its inputs vary. Taking the derivative of a function, whether scalar or vector, tells you the rate of change of the output value(s) with respect to an input parameter. For a vector function like \( r(t) = e^{-2t} i + \cos t \mathbf{j} + 3 \sin t \mathbf{k} \), the derivative \( \frac{dr}{dt} \) represents the tangent vector, showing both the direction and speed of change along the curve.

Working with Vector Derivatives

To find the derivative of \( r(t) \):

• Calculate the derivative of each component (\( -2e^{-2t}, -\sin t, 3\cos t \)).
• The result forms a new vector that gives the direction of the tangent to the curve at any \( t \).

Evaluating this against specific \( t \), such as \( t=0 \) in this exercise, determines the exact tangent vector at that point, crucial for identifying tangent lines.

An intersection with a plane involves finding where and how a line crosses a specified plane in space. In tasks involving vector lines and planes, like this problem, it's about locating where the tangent line meets another geometric space, here the y-plane.

• The y-plane refers to the set of all points where the x-coordinate is zero.
• We parameterized the equation of the tangent line (\( x = 1 - 2s, y = 1, z = 3s \)), which is derived once the tangent direction is known.
• To find the intersection with the y-plane, set \( x = 0 \).

Solving where \( 1 - 2s = 0 \) gives the parameter \( s \) that, when substituted back into the y and z equations, yields the intersection coordinates \( (0, 1, \frac{3}{2}) \). This exercise showcases a practical vector calculus application in analyzing space intersections.