Problem 45 - Calculus Early Transcendentals

Question

Find equations of the normal plane and osculating plane of the curve at the given point. \(x=2 \sin 3 t, y=t, z=2 \cos 3 t ; \quad(0, \pi,-2)\)

Step-by-Step Solution

Answer

Normal plane: \(-6x + y = \pi\); Osculating plane: \(-18x - 108y = -108\pi\).

Step 1: Parametrize the curve

The curve is given in parametric form: \( x = 2 \sin(3t) \), \( y = t \), and \( z = 2 \cos(3t) \). Identify the parameter \( t \) for the point \((0, \pi, -2)\). Based on the equation \(y = t\), set \( t = \pi \) to find the point on the curve.

Step 2: Calculate the tangent vector

Differentiate each component of the curve with respect to \(t\). This gives the tangent vector \( \mathbf{r}'(t) = (\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}) \). Thus, \( \mathbf{r}'(t) = (6 \cos(3t), 1, -6 \sin(3t)) \). Evaluate at \( t = \pi \): \( \mathbf{r}'(\pi) = (6 \cos(3\pi), 1, -6 \sin(3\pi)) = (-6, 1, 0) \).

Step 3: Calculate the normal vector

The normal vector to the curve is obtained by differentiating the tangent vector: \( \mathbf{r}''(t) = (-18 \sin(3t), 0, -18 \cos(3t)) \). Evaluate \( \mathbf{r}''(\pi) = (-18 \sin(3\pi), 0, -18 \cos(3\pi)) = (0, 0, -18) \).

Step 4: Determine the binormal vector

Calculate the binormal vector as the cross product of the tangent vector and the normal vector: \( \mathbf{B}(t) = \mathbf{r}'(t) \times \mathbf{r}''(t) \). Perform the cross product operation \((-6, 1, 0) \times (0, 0, -18) = (-18, -108, 0)\).

Step 5: Equation of the normal plane

The normal plane is perpendicular to the tangent vector. Use the tangent vector equation \((-6, 1, 0)\) and the given point \((0, \pi, -2)\) to form: \(-6(x - 0) + 1(y - \pi) + 0(z + 2) = 0 \Rightarrow -6x + y - \pi = 0\).

Step 6: Equation of the osculating plane

The osculating plane contains the tangent vector and the normal plane. Using the binormal vector \((-18, -108, 0)\), the equation is: \(-18(x - 0) - 108(y - \pi) + 0(z + 2) = 0\). Simplify to: \(-18x - 108y + 108\pi = 0\).

Key Concepts

Tangent Vector

A vital part in understanding curves is the tangent vector. Imagine you're drawing a curve on paper at any point along this curve. The tangent vector shows the direction in which the curve is heading at that point. To find it mathematically, we differentiate each of the parametric equations which describe the curve. So, if your curve is described by equations like:

• \(x = f(t)\)
• \(y = g(t)\)
• \(z = h(t)\)

Differentiating these with respect to \(t\) gives us derivatives \(\frac{dx}{dt}\), \(\frac{dy}{dt}\), and \(\frac{dz}{dt}\).
Together, these derivatives make up the tangent vector: \(\mathbf{r}'(t) = (\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt})\). For example, from the step-by-step solution, when plugging in \(t = \pi\), the tangent vector becomes \((-6, 1, 0)\).
This vector tells us that at the point \((0, \pi, -2)\), the curve moves primarily in the negative \(x\) direction, slightly in the \(y\) direction, and not at all in the \(z\) direction.

Parametric Equations

Parametric equations are a useful way to describe curves. Instead of describing a curve with one equation (like \(y = f(x)\)), we use a set of parametric equations, each dependent on a parameter \(t\). For example, the set of equations

• \(x = 2 \sin(3t)\)
• \(y = t\)
• \(z = 2 \cos(3t)\)

These equations describe a 3-dimensional curve where \(x\), \(y\), and \(z\) are each calculated from the same variable \(t\).
By adjusting \(t\), we can find any point along the curve. For a specific point like \((0, \pi, -2)\), we solve the equations to determine the appropriate \(t\).
We found that \(t = \pi\), because substituting \(t = \pi\) into \(y = t\) conveniently matches the \(y\) coordinate of the point. Parametric equations are powerful because they can easily describe complex curves and allow for straightforward calculations of derivatives and integrals.

Cross Product

The cross product is a vector operation useful in three-dimensional geometry. It calculates a vector that is perpendicular to two given vectors. If you have two vectors:

• \( \mathbf{a} = (a_1, a_2, a_3) \)
• \( \mathbf{b} = (b_1, b_2, b_3) \)

The cross product \( \mathbf{a} \times \mathbf{b} \) is given by:\[(a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\]
In the context of our problem, the cross product finds the binormal vector of the curve. The binormal vector, in turn, is useful for finding the osculating plane.
Here, the tangent vector is \((-6, 1, 0)\), and the normal vector derived from it is \((0, 0, -18)\).
Their cross product is \((-18, -108, 0)\). This result shows us the vector that's perpendicular to both the tangent and normal vectors. This perpendicularity is central in finding the equation of the osculating plane.