Question

Find the equation of the plane perpendicular to the curve $$ \mathbf{r}(t)=t \sin t \mathbf{i}+3 t \mathbf{j}+2 t \cos t \mathbf{k} $$ at \(t=\pi / 2\).

Short answer

The plane equation is \( 3y - \frac{\pi}{2}z = \frac{9\pi}{2} \).

Step-by-step solution

Identify the Tangent Vector

To find the normal to the plane, we first need the tangent vector to the given curve at the point where \( t = \pi/2 \). This tangent vector is the derivative of \( \mathbf{r}(t) \): \( \mathbf{r}'(t) = \frac{d}{dt}(t \sin t \mathbf{i} + 3t \mathbf{j} + 2t \cos t \mathbf{k}) \).

Compute Derivative

Calculate the derivative \( \mathbf{r}'(t) \) using the product rule: \( \mathbf{r}'(t) = (\sin t + t \cos t) \mathbf{i} + 3 \mathbf{j} + (2 \cos t - 2t \sin t) \mathbf{k} \).

Evaluate at \( t = \pi/2 \)

Substitute \( t = \pi/2 \) into the derivative to find the tangent vector at this point: \( \mathbf{r}'(\pi/2) = (1 \cdot 0 + \frac{\pi}{2} \cdot 0) \mathbf{i} + 3 \mathbf{j} + (2 \cdot 0 - \frac{\pi}{2} \cdot 1) \mathbf{k} = 0 \mathbf{i} + 3 \mathbf{j} - \frac{\pi}{2} \mathbf{k} \).

Find the Normal Vector

The normal vector to the plane is the same as the tangent vector at \( t=\pi/2 \), which is \( \mathbf{n} = 0 \mathbf{i} + 3 \mathbf{j} - \frac{\pi}{2} \mathbf{k} \).

Find the Point on the Plane

Substitute \( t = \pi/2 \) into the original vector function \( \mathbf{r}(t) \) to find a specific point on the plane: \( \mathbf{r}(\pi/2) = \frac{\pi}{2} \cdot 1 \mathbf{i} + 3 \cdot \frac{\pi}{2} \mathbf{j} + 2 \cdot \frac{\pi}{2} \cdot 0 \mathbf{k} = \frac{\pi}{2} \mathbf{i} + \frac{3\pi}{2} \mathbf{j} + 0 \mathbf{k} \).

Write the Equation of the Plane

The equation of a plane with normal vector \( \mathbf{n} = 0 \mathbf{i} + 3 \mathbf{j} - \frac{\pi}{2} \mathbf{k} \) through point \( (\frac{\pi}{2}, \frac{3\pi}{2}, 0) \) is \( 0(x - \frac{\pi}{2}) + 3(y - \frac{3\pi}{2}) - \frac{\pi}{2}(z - 0) = 0 \). Simplifying, \( 3y - \frac{\pi}{2}z = \frac{9\pi}{2} \).

Key concepts

Tangent Vector

When we study curves in space, the tangent vector is vital for understanding their direction at any given point. Imagine driving on a road; the direction of your car at any point is similar to a tangent vector of that road. In mathematical terms, a tangent vector to a curve is derived from the curve's function using derivatives. It's like capturing the instantaneous velocity of the curve at a specific point.

The tangent vector holds the direction in which the curve proceeds. To find it, you'd perform the derivative of the curve's parametric equations. For the curve given, \( \mathbf{r}(t) = t \sin t \mathbf{i} + 3t \mathbf{j} + 2t \cos t \mathbf{k} \), its derivative gives the tangent vector \( \mathbf{r}'(t) \). This derivative provides the rate of change of the curve in the \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) directions, helping us visualize how the curve moves through space.

Normal Vector

A normal vector stands out as it is perpendicular to the tangent vector. Imagine a plane touching a basketball at one single point; the line going straight up from that point would be like the normal vector. It's key in things like defining a plane which can be understood as a flat surface through this perpendicular relationship.

In addition to determining how a plane or a surface bends, normal vectors also help in identifying perpendicular directions in three-dimensional space. In step 4 of the solution, we found that the normal vector for the plane is also the tangent vector \( \mathbf{n} = 0 \mathbf{i} + 3 \mathbf{j} - \frac{\pi}{2} \mathbf{k} \) computed at the point \( t = \pi/2 \). This vector becomes the anchor to the plane, aiding in linking various geometric concepts like angles and perpendicular lines.

Calculus

Calculus is the mathematical study that helps us understand change—whether it's speed in physics or trends in data. By employing calculus, you can determine where a curve is heading, how fast it's changing, and where it was or will be.

In problems dealing with curves, calculus allows the calculation of derivatives which are crucial in determining tangent vectors through differentiation. This step-by-step differentiation of functions forms the core of calculus. It's essential when we need precise measurements of how a function behaves at particular points, as seen with finding the tangent and normal vectors in the example problem.

• Calculating the rate of change
• Understanding direction via tangent vectors
• Utilizing derivatives for exact measurements

These concepts are fundamental to solving plane equations, optimizing functions, and exploring deeper mathematical territories.

Derivative

The derivative is a cornerstone concept in calculus, representing how a function changes as its input changes. Think of the stock market graph—its sharp climbs and plummeting descents can be analyzed using derivatives to quantify the rate of change.

Derivatives allow us to peel back the layers of curves and surfaces to reveal their true nature. In mathematics, the derivative of a function gives you the slope of the tangent line to the graph of the function at any given point. Upon calculating the derivative of \( \mathbf{r}(t) \), we obtained \( \mathbf{r}'(t) = (\sin t + t \cos t) \mathbf{i} + 3 \mathbf{j} + (2 \cos t - 2t \sin t) \mathbf{k} \).

When you evaluate this expression at a specific value like \( t = \pi/2 \), it translates to calculating the precise direction the point on the curve is moving towards. Derivatives, hence, offer us an essential tool to probe into curves, giving us the tangent vector direction and enabling us to define planes effectively.