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Chapter 14: Problem 17

Normal and tangential components Determine the points (if any) on the curve C at which the vector field \(\mathbf{F}\) is tangent to C and normal to C. Sketch C and a few representative vectors of \(\mathbf{F}\). $$\mathbf{F}=\langle x, y\rangle, \text { where } C=\left\\{(x, y): x^{2}+y^{2}=4\right\\}$$

Short Answer

Expert verified
The vector field \(\mathbf{F}=\langle x, y\rangle\) is both tangent and normal to the curve \(C: x^2+y^2=4\) at all points. The points on the curve \(C\) can be represented as \((2\cos(\theta), 2\sin(\theta))\) for any angle \(\theta\).

Step by step solution

01

Parametric representation of the curve C

To find the tangent and normal points, we need to express the curve C in terms of a parameter. In this case, the parameter will be the angle theta. The equation of the curve C is \(x^2+y^2=4\), which can be written as \(x=2\cos(\theta)\) and \(y=2\sin(\theta)\). So, the points on the curve C can be represented as \((2\cos(\theta), 2\sin(\theta))\).
02

Calculate tangent vectors at various points on the curve C

Now that we have the parametric representation of the curve, we need to find the tangent vectors at various points on the curve. We can do this by finding the derivative of the parametric equations with respect to the parameter theta: \(\frac{dx}{d\theta} = -2\sin(\theta)\) and \(\frac{dy}{d\theta} = 2\cos(\theta)\). The tangent vector at any point on the curve C will then be \(\mathbf{T}(\theta) =\langle -2\sin(\theta), 2\cos(\theta)\rangle\).
03

Evaluate the dot product and cross product between the vector field and tangent vectors

Now, we will evaluate the dot product between the vector field \(\mathbf{F}\) and the tangent vector \(\mathbf{T}(\theta)\): \(\mathbf{F}\cdot\mathbf{T}(\theta) = (x)(-2\sin(\theta)) + (y)(2\cos(\theta))\). Since \(x=2\cos(\theta)\) and \(y=2\sin(\theta)\), we can substitute these values in the dot product: \(\mathbf{F}\cdot\mathbf{T}(\theta) = (2\cos(\theta))(-2\sin(\theta)) + (2\sin(\theta))(2\cos(\theta)) = 0\). Since the dot product is zero for all values of theta, the vector field \(\mathbf{F}\) is normal to the curve C at all points. Next, we evaluate the cross product magnitude between the vector field \(\mathbf{F}\) and the tangent vector \(\mathbf{T}(\theta)\) in 2D: \(||\mathbf{F}\times\mathbf{T}(\theta)|| = ||\mathbf{F}||||\mathbf{T}(\theta)||\sin(\phi)\), where \(\phi\) is the angle between the vector field and the tangent vector. By computing the magnitudes of the vector field and the tangent vector, we find: \(||\mathbf{F}|| =\sqrt{x^2+y^2} = 2\) and \(||\mathbf{T}(\theta)|| = \sqrt{(-2\sin(\theta))^2 + (2\cos(\theta))^2} = 2\). Since the dot product is zero, the angle \(\phi\) will be either \(0\) or \(\pi\). Therefore, the cross product magnitude will be zero, and the vector field \(\mathbf{F}\) is tangent to the curve C at all points. In conclusion, the vector field \(\mathbf{F}\) is both normal and tangent to the curve C at all points.

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Key Concepts

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

Parametric Representation
When dealing with curves like curve C given by the equation \(x^2 + y^2 = 4\), it's often helpful to use a parametric representation. This means expressing both \(x\) and \(y\) in terms of another variable, typically an angle \(\theta\) in this case. We use trigonometric identities for this transformation:

\[ x = 2\cos(\theta), \quad y = 2\sin(\theta) \]

By doing this, each point on the curve can be clearly described as a pair \((x, y)\) where \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\). This representation helps when finding other properties such as tangents and normals to the curve.
  • \(x^2 + y^2\) becomes \((2\cos(\theta))^2 + (2\sin(\theta))^2 = 4\).
  • This method uses the natural cyclic nature of trigonometric functions, making calculations straightforward using angles.
Tangent Vector
Finding a tangent vector to a curve is essential to understand the direction of the curve at any given point. For the curve C, after representing \(x\) and \(y\) parametric terms of \(\theta\), the tangent vector \(\mathbf{T}(\theta)\) is found by differentiating these parametric equations with respect to \(\theta\).

Thus, the tangent vector is given by:
\[ \mathbf{T}(\theta) = \left\langle \frac{dx}{d\theta}, \frac{dy}{d\theta} \right\rangle = \left\langle -2\sin(\theta), 2\cos(\theta) \right\rangle \]
  • \(\frac{dx}{d\theta}\) results from differentiating \(2\cos(\theta)\), which is \(-2\sin(\theta)\).
  • \(\frac{dy}{d\theta}\) results from differentiating \(2\sin(\theta)\), leading to \(2\cos(\theta)\).
This vector describes how rapidly and in which direction a point on the curve is moving as \(\theta\) changes.
Vector Field Dot Product
The dot product is a calculation that tells us how much of one vector goes in the direction of another. For the vector field \(\mathbf{F} = \langle x, y \rangle\) and the tangent vector \(\mathbf{T}(\theta)\) of curve C, the dot product is a key tool to determine if these vectors are normal, or perpendicular, at various points.

We calculate:
\[\mathbf{F} \cdot \mathbf{T}(\theta) = (x)(-2\sin(\theta)) + (y)(2\cos(\theta))\]
By inserting parameterized versions \(x = 2\cos(\theta)\) and \(y = 2\sin(\theta)\), the complex simplify:
\[\mathbf{F} \cdot \mathbf{T}(\theta) = (2\cos(\theta))(-2\sin(\theta)) + (2\sin(\theta))(2\cos(\theta)) = 0\]
  • A dot product of zero indicates the vector field is normal (perpendicular) to the curve.
  • Since this is true for all \(\theta\), \(\mathbf{F}\) is consistently normal to the curve at every point.
Cross Product Magnitude
For 2D vectors, while a true cross product doesn't strictly exist as it does in 3D, we can still consider magnitudes that capture perpendicularity and parallelism. Here, the cross product magnitude helps determine that the vector field is tangent to the curve C.

The cross product magnitude in 2D is expressed as:
\[||\mathbf{F} \times \mathbf{T}(\theta)|| = ||\mathbf{F}||||\mathbf{T}(\theta)||\sin(\phi)\]
Where:
  • \(\mathbf{F} = \langle x, y \rangle\)
  • \(||\mathbf{F}|| = \sqrt{x^2 + y^2} = 2\) for the circle \(x^2 + y^2 = 4\)
  • \(||\mathbf{T}(\theta)|| = 2\)
Given that the dot product is \(0\), the angle \(\phi\) between \(\mathbf{F}\) and \(\mathbf{T}(\theta)\) is \(\frac{\pi}{2}\), resulting in a cross product magnitude of zero. This tells us that, in addition to being normal, the vector field can also be considered tangent at every point due to these zero-cross magnitude conditions.

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Most popular questions from this chapter

\(A\) scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\) Show that if \(\varphi\) is harmonic on a region \(D\) enclosed by a surface \(S\) then $$\iint_{S} \nabla \varphi \cdot \mathbf{n} d S=0$$

The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k>0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\). \(T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; S\) is the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\).

Show that \(\left|\mathbf{t}_{u} \times \mathbf{t}_{v}\right|=a^{2} \sin u\) for a sphere of radius \(a\) defined parametrically by \(\mathbf{r}(u, v)=\langle a \sin u \cos v, a \sin u \sin v, a \cos u\rangle,\) where \(0 \leq u \leq \pi\) and \(0 \leq v \leq 2 \pi\).

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Prove the following identities. Assume that \(\varphi\) is \(a\) differentiable scalar-valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \times(\mathbf{F} \times \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}-\mathbf{G}(\nabla \cdot \mathbf{F})-(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F}(\nabla \cdot \mathbf{G})$$
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