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Consider the curve \(\mathbf{r}(t)=2 t \mathbf{i}+t^{2} \mathbf{j}+\left(1-t^{2}\right) \mathbf{k}\) (a) Show that this curve lies on a plane and find the equation of this plane. (b) Where does the tangent line at \(t=2\) intersect the \(x y\) -plane?

Short Answer

Expert verified
(a) The plane equation is \( z + y = 1 \). (b) Intersection at (2.5, 1, 0).

Step by step solution

01

Find the Plane Equation

To show that the curve lies on a plane, express the curve components in terms of one another. The given parametric equations are: 1. \( x(t) = 2t \)2. \( y(t) = t^2 \)3. \( z(t) = 1 - t^2 \)We can relate \( z \) and \( y \) directly: \( z = 1 - y \). Therefore, the equation of the plane is \( z + y = 1 \). This shows the curve lies on the plane defined by this equation.
02

Calculate the Tangent Line at t=2

First, find \( \mathbf{r}'(t) \), the derivative of \( \mathbf{r}(t) \), which gives the direction of the tangent line. 1. The derivative is \( \mathbf{r}'(t) = 2\mathbf{i} + 2t\mathbf{j} - 2t\mathbf{k} \).At \( t = 2 \), this simplifies to \( \mathbf{r}'(2) = 2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k} \). The point on the curve at \( t = 2 \) is \( \mathbf{r}(2) = 4\mathbf{i} + 4\mathbf{j} - 3\mathbf{k} \). Using this point and direction vector, the tangent line is: \[ \mathbf{L}(s) = (4 + 2s)\mathbf{i} + (4 + 4s)\mathbf{j} + (-3 - 4s)\mathbf{k} \]
03

Find Intersection with the XY-plane

For the tangent line to intersect the \( xy \)-plane, \( z \) must be \( 0 \). Using the tangent line equation: \[ -3 - 4s = 0 \] Solving for \( s \), we find \( s = -\frac{3}{4} \). Substituting \( s = -\frac{3}{4} \) into the expressions for \( x(s) \) and \( y(s) \): \( x = 4 + 2(-\frac{3}{4}) = 2.5 \) \( y = 4 + 4(-\frac{3}{4}) = 1 \)Thus, the intersection point is \( (2.5, 1, 0) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation of Plane
When dealing with parametric curves, determining if they lie on a plane is a common problem. For the curve \( \mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + (1 - t^2) \mathbf{k} \), we can express each component \( x(t) \), \( y(t) \), and \( z(t) \) to reveal the relation between them. Here:
  • \( x(t) = 2t \)
  • \( y(t) = t^2 \)
  • \( z(t) = 1 - t^2 \)
Notice that both \( y(t) \) and \( z(t) \) depend on \( t^2 \), allowing us to relate them directly: \( z + y = 1 \). This equation \( z + y = 1 \) defines a plane in three-dimensional space showing that at any value of \( t \), the curve lies on this plane.
Tangent Line
The tangent line of a parametric curve at a specific point provides insights into its behavior at that point. To find the tangent line to our curve \( \mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + (1 - t^2) \mathbf{k} \) at \( t = 2 \), we begin by calculating its derivative, \( \mathbf{r}'(t) \). This derivative gives the direction of the tangent to the curve:
  • \( \mathbf{r}'(t) = 2\mathbf{i} + 2t\mathbf{j} - 2t\mathbf{k} \)
Plugging \( t = 2 \) into \( \mathbf{r}'(t) \) yields \( \mathbf{r}'(2) = 2\mathbf{i} + 4\mathbf{j} - 4\mathbf{k} \). The position at this point is \( \mathbf{r}(2) = 4\mathbf{i} + 4\mathbf{j} - 3\mathbf{k} \). Therefore, the tangent line \( \mathbf{L}(s) \) combining this direction and position can be expressed as:\[ \mathbf{L}(s) = (4 + 2s)\mathbf{i} + (4 + 4s)\mathbf{j} + (-3 - 4s)\mathbf{k} \].
Intersection with XY-plane
Finding where the tangent line intersects with the \( xy \)-plane involves setting its \( z \)-coordinate to 0. This is because the \( xy \)-plane is described by \( z = 0 \). For our tangent line \( \mathbf{L}(s) = (4 + 2s)\mathbf{i} + (4 + 4s)\mathbf{j} + (-3 - 4s)\mathbf{k} \), we solve:
  • \(-3 - 4s = 0 \)
Solving this equation, we find \( s = -\frac{3}{4} \). Substituting this value back into \( x \) and \( y \) gives:
  • \( x = 4 + 2(-\frac{3}{4}) = 2.5 \)
  • \( y = 4 + 4(-\frac{3}{4}) = 1 \)
Thus, the point of intersection with the \( xy \)-plane is \( (2.5, 1, 0) \), which provides valuable information about where this tangent line reaches the plane.

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