For the vector field 𝐅 and curve C, complete the following: a. Determine the points (if any) alo

For the vector field 𝐅 and curve C, complete the following:...

For the vector field $\mathbf{F}$ and curve $C$, complete the following: a. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is tangent to $C$. b. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is normal to $C$ c. Sketch $C$ and a few representative vectors of $\mathbf{F}$ on $C$. $$\mathbf{F}=\left\langle\frac{y}{2},-\frac{x}{2}\right\rangle ; C=\left\{(x, y): y-x^{2}=1\right\}$$

For the vector field $\mathbf{F}$ and curve $C$, complete the following:
a. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is tangent to $C$.
b. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is normal to $C$
c. Sketch $C$ and a few representative vectors of $\mathbf{F}$ on $C$.

$$\mathbf{F}=\left\langle\frac{y}{2},-\frac{x}{2}\right\rangle ; C=\left\{(x, y): y-x^{2}=1\right\}$$

Key Concepts

Vector Field

A vector field is a mapping that assigns a vector to every point in a space, such as the plane. It is used to represent varying directional quantities like velocity or force and is fundamental in exploring how these quantities interact with curves or surfaces within the field.

Implicitly Defined Curve

An implicitly defined curve is described by an equation relating the coordinates (for example, an equation like y - x^2 = 1). Such curves often require techniques like implicit differentiation to analyze their geometric properties, including slopes and tangent directions.

Tangent Vector to a Curve

The tangent vector at a point on a curve indicates the direction in which the curve is heading at that point. It can be determined by differentiating the curve's defining equation, and by comparing it with a vector field, one can determine if the field is tangent to the curve when the two vectors are parallel.

Normal Vector

A normal vector to a curve is a vector that is perpendicular to the tangent vector at a given point. Normal vectors are key in applications such as computing flux, understanding gradients, or analyzing forces acting perpendicularly to the curve.

Dot Product and Orthogonality

The dot product is a mathematical operation that, when applied to two vectors, produces a scalar value. It is essential for checking orthogonality because if the dot product of two vectors is zero, the vectors are perpendicular. This concept is used to determine when a vector field is normal to a curve.

Transcript

00:01 This question gives us a specific vector field and a specific curve and asks us to find a few things from this.

00:06 We need to know what points, if any, where our vector field is tangent to the curve, what points it's normal to the curve, and then we're going to want to sketch our curve and then list some of the vectors f on the curve.

00:22 So our vector f is y over 2, comma, negative x over 2.

00:27 Our curve c is shown by the curve created from y minus x squared is equal to one.

00:33 Before we do anything, we need to parameterize this curve.

00:36 Well, by adding x squared to both sides, we have y is equal to x squared plus one.

00:41 Because we have an explicit formula for y, we can just do this.

00:45 We can make it xt and then y would be t squared plus one, which satisfies that equation that we have there.

00:53 It makes sense.

00:54 So now we can get on with our parts.

00:57 Was to find the points where f is tangent to c.

01:00 So this means points where f and our tangent curve are the same.

01:06 How do we find a tangent curve? we take the derivative of both components with respect to t.

01:11 So we have 1 comma 2t, just taking the derivative there.

01:15 And so now what we need to do is we need to see any points for which y over 2 comma x over 2.

01:24 Dot 1 comma 2t is equal to 1 since this would mean they're parallel so that it would impact to be the same vector so doing the stop product we have y over 2 times 1 plus negative x over 2 times 2 t is equal to 1 so we have a problem we have a mixture of x's and y's and t's well this can be solved by using the formula we have here so x is t y is t squared plus one since we're only looking at the points that are on c this would make sense so we have t squared plus one over two times one plus negative t over two times two t is equal to one further we get t squared plus one over two plus negative two t squared over two equals one so combining these, we would get t squared plus 1 minus 2 t squared over 2 is equal to 1.

02:34 Or we can go ahead and multiply both sides by 2 and combine these terms.

02:39 We have 1 minus t squared is equal to 2 or 1 is equal to 2 plus t squared.

02:48 And that would be t squared is equal to negative 1.

02:52 That doesn't have any real solutions...

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