Question

Let the position vector (with its tail at the origin) of a moving particle be \(\mathbf{r}=\mathbf{r}(t)=\) \(t^{2} \mathbf{i}-2 t \mathbf{j}+\left(t^{2}+2 t\right) \mathbf{k},\) where \(t\) represents time. (a) Show that the particle goes through the point (4,-4,8) . At what time does it do this? (b) Find the velocity vector and the speed of the particle at time \(t ;\) at the time when it passes though the point (4,-4,8). (c) Find the equations of the line tangent to the curve described by the particle and the plane normal to this curve, at the point (4,-4,8).

Short answer

At \( t = 2 \). Velocity vector is \( 2t \mathbf{i} - 2 \mathbf{j} + (2t + 2) \mathbf{k} \). Speed is \( 2\sqrt{2(t+1)} \). Tangent line and normal plane equations can be derived from velocity vector.

Step-by-step solution

Verify the Position Vector at a Point

Substitute the coordinates of the point (4, -4, 8) into the position vector equation to find the time \( t \). Check if the given coordinates satisfy the equation \( \mathbf{r}(t) = t^2 \mathbf{i} - 2t \mathbf{j} + (t^2 + 2t) \mathbf{k} \).

Solve for Time \( t \)

Set up the equations: \( t^2 = 4 \), \( -2t = -4 \), and \( t^2 + 2t = 8 \). Solve each equation for \( t \). All the equations should yield the same value of \( t \).

Find Velocity Vector \( \mathbf{v}(t) \)

Differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). \( \mathbf{v}(t) = \frac{d}{dt} [t^2 \mathbf{i} - 2t \mathbf{j} + (t^2 + 2t) \mathbf{k}] \)

Compute the Speed of the Particle

Find the magnitude of the velocity vector \( \mathbf{v}(t) \). The speed is \( \|\mathbf{v}(t)\| \). Substitute \( t \) obtained from Step 2 into this expression to get the speed at that specific time.

Determine the Tangent Line Equation

At the time found, use the position vector and velocity vector to write the parametric equations of the tangent line. The tangent line equations in terms of a parameter \( s \) are: \( x = x_0 + v_{x} s \), \( y = y_0 + v_{y} s \), \( z = z_0 + v_{z} s \).

Determine the Normal Plane Equation

Use the normal vector, which is the velocity vector \( \mathbf{v}(t) \), at the point (4, -4, 8) to write the equation of the normal plane. The plane equation takes the form \(v_x(x - x_0) + v_y(y - y_0) + v_z(z - z_0) = 0\).

Key concepts

Position Vector

To understand the movement of a particle in 3D space, we use a position vector. The position vector, denoted as \(\textbf{r}(t)\), describes the coordinates of a particle at any time \( t \). In the exercise, the position vector is given by\(\textbf{r}(t) = t^2 \textbf{i} - 2t \textbf{j} + (t^2 + 2t) \textbf{k} \). This means at any time \( t \), you can plug it into this equation to find the exact location of the particle. For example, at time \( t = 2 \), the coordinates will be \( (2^2, -2 \times 2, 2^2 + 2 \times 2) = (4, -4, 8) \). The exercise asks us to verify if the particle goes through the point (4, -4, 8). By solving the equations derived from the position vector at this point, we find \( t \). This is crucial in helping us understand when the particle is at a certain position in space.

Velocity Vector

The velocity vector describes how fast and in what direction the particle is moving at any time \( t \). It is found by differentiating the position vector with respect to time. For our exercise, we start with \(\textbf{r}(t) = t^2 \textbf{i} - 2t \textbf{j} + (t^2 + 2t) \textbf{k} \) and find its derivative \(\textbf{v}(t)\):

\( \textbf{v}(t) = \frac{d}{dt}[t^2 \textbf{i} - 2t \textbf{j} + (t^2 + 2t) \textbf{k}] = 2t \textbf{i} - 2 \textbf{j} + (2t + 2) \textbf{k} \).

This gives us the velocity vector at any time \( t \). When the particle passes through the point (4,-4,8), substituting the value \( t = 2 \) into the velocity vector we get \(\textbf{v}(2) = 4 \textbf{i} - 2 \textbf{j} + 6 \textbf{k} \). This indicates that, at \( t = 2 \), the particle is moving with components 4 in the x-direction, -2 in the y-direction, and 6 in the z-direction.

Tangent Line

The tangent line to a curve at a given point shows the direction in which the curve is heading at that point. To find the tangent line at the point where \(\textbf{r}(2) = (4, -4, 8)\), we use both the position vector and the velocity vector at that time. The equations for the tangent line are derived from:

\( x = x_0 + v_x s \)
\( y = y_0 + v_y s \)
\( z = z_0 + v_z s \),

where \( v_x, v_y, \texbf{d} v_z \) are the components of the velocity vector, and \texbf{\textbf{r}(0)} are the coordinates of the point (4, -4, 8). Substituting the values we get:

\( x = 4 + 4s \)
\( y = -4 - 2s \)
\( z = 8 + 6s \),

where \( s \) is a parameter describing the line. This equation tells us how the curve behaves locally at the point (4, -4, 8).

Normal Plane

The normal plane to a curve at a given point is perpendicular to the tangent line at that point. To find the equation of the normal plane at the point (4, -4, 8), we use the velocity vector \( \textbf{v}(2) = 4 \textbf{i} - 2 \textbf{j} + 6 \textbf{k} \) as the normal vector to the plane.

The equation of the normal plane takes the form:

\( v_x (x - x_0) + v_y (y - y_0) + v_z (z - z_0) = 0 \).

Substituting the values, we get:

\[4 (x - 4) - 2 (y + 4) + 6 (z - 8) = 0 \],

which simplifies to

\[4x - 2y + 6z - 60 = 0 \].

This equation describes a plane orthogonal to the curve at the point (4,-4,8).