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Chapter 12: Problem 6

In what plane does the curve \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{k}\) lie?

Short Answer

Expert verified
Question: Determine the equation of the plane containing the curve \(\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{k}\). Answer: The equation of the plane containing the curve is \(y = 0\).

Step by step solution

01

Extract directional vectors from the curve equation

We are given the curve \(\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{k}\). Let's parameterize the curve as follows: \(x = t\), \(y = 0\) (since there is no \(\mathbf{j}\) component), and \(z = t^2\). Now, let's find a vector \(\vec{V}\) tangent to the curve. To do this, we will compute the derivative of \(\mathbf{r}(t)\) with respect to \(t\): \(\frac{d\mathbf{r}}{dt} = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j} + \frac{dz}{dt}\mathbf{k}\).
02

Compute the derivative of \(\mathbf{r}(t)\)

Compute the derivative for each component: \(\frac{dx}{dt} = 1\), \(\frac{dy}{dt} = 0\), and \(\frac{dz}{dt} = 2t\). Thus, \(\frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t\mathbf{k}\). Since this tangent vector lies in the plane, we can take this as one of the directional vectors in the plane, denoted by \(\vec{V}\). Now, \(\vec{V} = \mathbf{i} + 2t\mathbf{k}\).
03

Determine a second tangent vector in the plane

Let's find another vector tangent to the curve. We can take the derivative of \(\frac{d\mathbf{r}}{dt}\) with respect to \(t\) to find a second direction vector that lies in the plane: \(\frac{d^2\mathbf{r}}{dt^2} = \frac{d^2x}{dt^2}\mathbf{i} + \frac{d^2y}{dt^2}\mathbf{j} + \frac{d^2z}{dt^2}\mathbf{k}\). Compute the second derivative for each component: \(\frac{d^2x}{dt^2} = 0\), \(\frac{d^2y}{dt^2} = 0\), and \(\frac{d^2z}{dt^2} = 2\). Thus, \(\frac{d^2\mathbf{r}}{dt^2} = 2\mathbf{k}\). We can denote this second directional vector by \(\vec{W}\), so \(\vec{W} = 2\mathbf{k}\).
04

Find a point on the curve

To find a point on the curve, we plug in any value for \(t\) into the parametric equation. We will choose \(t=0\): \(x = 0\), \(y = 0\), and \(z = 0^2 = 0\). Thus, we have the point \(P = (0, 0, 0)\) on the curve.
05

Compute the normal vector to the plane

We can find a vector normal to the plane by taking the cross product of two directional vectors in the plane: \(\vec{N} = \vec{V} \times \vec{W}\). \(\vec{N} = (\mathbf{i} + 2t\mathbf{k}) \times (2\mathbf{k}) = 2\mathbf{i} \times 2\mathbf{k} = 4\mathbf{j}\).
06

Find the equation of the plane

Using the point-normal form of the plane equation, we have: \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\), where \((A, B, C)\) are the components of the normal vector \(\vec{N}\), and \((x_0, y_0, z_0)\) are the coordinates of the point \(P\). Plug in the values: \(4(y - 0) = 0 \Rightarrow 4y = 0\). The equation of the plane containing the curve is \(y = 0\).

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