Problem 22 A curve $C$ is given parametri... [FREE SOLUTION]

Chapter 15: Problem 22

A curve $C$ is given parametrically. Find two unit tangent vectors to $C$ at $P$. $$ x=\sin t+2, \quad y=\cos t, \quad z=t ; \quad P(2,1,0) $$

Short Answer

Expert verified

The two unit tangent vectors at $ P $ are $ \langle \frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \rangle $ and $ \langle -\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}} \rangle $.

Step by step solution

Verify the Point on the Curve

First, ensure the given point $ P(2,1,0) $ lies on the curve. Use the parametric equations $ x = \sin t + 2 $, $ y = \cos t $, and $ z = t $. Apply the conditions of the point to these equations: $ \sin t + 2 = 2 $, $ \cos t = 1 $, and $ t = 0 $. Solving these gives $ \sin t = 0 $ and $ \cos t = 1 $, which are satisfied for $ t = 0 $. Therefore, $ P(2,1,0) $ is on the curve at $ t = 0 $.

Find the Derivative of Each Component

Determine the tangent vector by differentiating the parametric equations with respect to $ t $. Compute $ \frac{dx}{dt} = \cos t $, $ \frac{dy}{dt} = -\sin t $, and $ \frac{dz}{dt} = 1 $. At $ t = 0 $, these become $ \frac{dx}{dt} = 1 $, $ \frac{dy}{dt} = 0 $, and $ \frac{dz}{dt} = 1 $. Thus, the tangent vector at $ t = 0 $ is $ \langle 1, 0, 1 \rangle $.

Normalize the Tangent Vector

To find a unit tangent vector, normalize the tangent vector $ \langle 1, 0, 1 \rangle $. Calculate the magnitude of this vector: $ \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2} $. Normalize by dividing each component by $ \sqrt{2} $, giving $ \langle \frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \rangle $.

Determine the Two Unit Tangent Vectors

The tangent vector found can have both a positive and a negative direction. Therefore, the two unit tangent vectors at $ P $ are $ \langle \frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \rangle $ and $ \langle -\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}} \rangle $.

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Key Concepts

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

Understanding the Unit Tangent Vector

A unit tangent vector to a curve represents a direction along the curve at a specific point. It's crucial in understanding how the curve behaves at that point.
The tangent vector itself shows the direction of the curve. To convert it into a unit tangent vector, it must have a length, or magnitude, of one.
This is done by normalizing the tangent vector.

First, compute the tangent vector by differentiating the parametric equations.
Next, calculate the magnitude of the tangent vector.
Finally, divide each component of the tangent vector by its magnitude.

This process ensures that the unit tangent vector points in the same direction but has a standardized length, which is valuable for calculations like curvature and directional derivatives.

Differentiation for Tangent Vectors

Differentiation is a fundamental technique in calculus critical for finding tangent vectors on parametric curves. To determine a tangent vector to a parametric curve, each component of the curve's parametric equations needs to be differentiated with respect to the parameter, often denoted as "t."

The x-component, derived as $ \frac{dx}{dt} $, gives the change rate of x with respect to t.
The y-component, $ \frac{dy}{dt} $, represents the change in y as t varies.
Similarly, the z-component, $ \frac{dz}{dt} $, shows the change in z.

Once differentiated, these components combine to form the tangent vector, $ \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle $. Evaluating these derivatives at a particular value of t gives the specific direction of the curve at that point.

Curve Verification Through Parametric Equations

Before diving deeper into calculations, it's essential to verify that a given point lies on the curve derived from parametric equations. This step ensures that all subsequent derivative and tangent calculations are at correct positions on the curve. To verify a point like $ P(2,1,0) $, plug it back into the original parametric equations:

If the x-component $ \sin t + 2 = 2 $ holds true, then $ \sin t = 0 $.
If $ \cos t = 1 $ works for the y-component, it confirms $ \cos t = 1 $.
The z-component checks if $ z = t $ is already satisfied with $ t = 0 $.

When these conditions are satisfied, it confirms the point is indeed part of the parameterized curve for the chosen parameter value.

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91影视

A curve \(C\) is given parametrically. Find two unit tangent vectors to \(C\) at \(P\). $$ x=\sin t+2, \quad y=\cos t, \quad z=t ; \quad P(2,1,0) $$

Short Answer

Step by step solution

Verify the Point on the Curve

Find the Derivative of Each Component

Normalize the Tangent Vector

Determine the Two Unit Tangent Vectors

Key Concepts

Understanding the Unit Tangent Vector

Differentiation for Tangent Vectors

Curve Verification Through Parametric Equations

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