(a) Find the point of intersection of the tangent lines to the curve 𝐫(t)=⟨sinπt, 2 sinπt, cosπt

(a) Find the point of intersection of the tangent lines to...

(a) Find the point of intersection of the tangent lines to the
curve $\mathbf{r}(t)=\langle\sin \pi t, 2 \sin \pi t, \cos \pi t\rangle$ at the points
where $ t=0$ and $t=0.5$ .
(b) Illustrate by graphing the curve and both tangent lines.

Key Concepts

Parametric Curves

Parametric curves represent geometric objects where the coordinates are defined as functions of a parameter rather than by a single equation. This method allows for the flexible description of curves and surfaces in multidimensional spaces and is especially useful in modeling complex paths and motions in physics and engineering.

Vector Derivatives

The derivative of a vector-valued function is obtained by differentiating each component with respect to the parameter, resulting in a vector that represents the rate of change or velocity of the original function. This derivative is fundamental in analyzing the behavior of curves, including computing directions for tangent lines.

Tangent Lines

A tangent line to a curve is a straight line that locally approximates the curve at a specific point; it passes through the point of tangency and has a direction given by the derivative at that point. In the context of parametric curves, the tangent line is computed using the point on the curve and the corresponding derivative vector.

Intersection of Lines

Finding the intersection of lines involves solving a system of equations where each line is expressed in a parametric form. The goal is to determine a common point in space that satisfies both line equations simultaneously, which is a key technique in both algebra and analytic geometry.

Graphing in 3D Space

Graphing in three-dimensional space involves plotting objects that extend in three dimensions, typically along the x, y, and z axes. This visualization helps in understanding spatial relationships and interactions between curves, surfaces, and their tangent lines, providing an intuitive grasp of the geometry involved.

Transcript

00:02 We are given the vector function for a curve and two points on this curve.

00:13 In part a, we are asked to find the point of intersection of the tangent line to this curve at these points.

00:24 The vector function is r of t equals sine of pi t, two sine of pi t, cosine pi t, and the two points are where t equals 0 and t equals 0 .5.

00:39 To find the tangent lines to the curve at these points, we want to find value of r of these points.

00:50 So in part a, we have that r of 0 is sine of 0, which is 0, 2 sine of 0, which is 0, and cosine of 0, which is 1.

01:10 And we have that r of 0 .5 is sine of pi over 2, which is 1, 2 sine of pi over 2, which is 2, and 2, and cosine of pi over 2, which is 0.

01:29 To find the tangent line will also need tangent vector, we have that the component functions of r, which are sine of pi t, 2 sine of pi t, and cosine of pi t are all differentiable.

01:48 So it follows that r is differentiable, and that r prime of t equals, by definition, the vector whose component functions are the derivatives of the component functions of r.

02:03 So this is pi cosine pi t, 2 pi cosine pi and negative pi sine pi t.

02:28 So the tangent vector at 0 is r prime of 0, which is pi cosine of 0, which is pi, 2 pi cosine of 0, which is 2 pi, and negative pi sine of 0, which is 0, and negative pi sine of 0, which is 0.

02:52 And the tangent vector when t equals 0 .5 is r prime of 0 .5, which is pi cosine of pi over 2, which is 0, 2 pi cosine of pi over 2, which is 0, and negative pi sign of pi over 2, which is negative pi.

03:30 So now we have both a point and a vector parallel to each tangent line.

03:46 So we can find the equations for both tangent lines...

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