Understanding Tangent and Gradient Vectors

Understanding Tangent and Gradient Vectors

Assessment

Interactive Video

Created by

Lucas Foster

Mathematics

10th - 12th Grade

Hard

The video tutorial explains how to find a unit tangent vector to a level curve at a given point. It begins by introducing the concept of tangent vectors and their relationship with gradient vectors, which are perpendicular to level curves. The tutorial then details the process of determining the gradient vector function and evaluating it at a specific point. It explains how to find a tangent vector by using the negative reciprocal of the gradient vector's slope. Finally, the tutorial demonstrates how to calculate a unit tangent vector with a positive X component by dividing the tangent vector by its magnitude.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main goal of the problem discussed in the video?

To solve a system of linear equations.

To calculate the area under a curve.

To find a unit tangent vector to a level curve at a specific point.

To determine the maximum value of a function.

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the relationship between the gradient vector and the level curve?

The gradient vector is tangent to the level curve.

The gradient vector is perpendicular to the level curve.

The gradient vector is equal to the level curve.

The gradient vector is parallel to the level curve.

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How can the tangent vector be determined using the gradient vector?

By finding the negative reciprocal of the gradient vector's slope.

By multiplying the gradient vector by a constant.

By adding the gradient vector to the level curve.

By subtracting the gradient vector from the level curve.

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in determining the gradient vector function?

Finding the maximum value of the function.

Solving the function for x and y.

Rewriting the function f(x, y).

Calculating the derivative of the function.

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the X component of the gradient vector at the point (1, 1)?

0

2

8/3

4

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the Y component of the gradient vector at the point (1, 1)?

8/3

0

4

2

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the slope of the tangent vector to the level curve at the point (1, 1)?

3/2

-3/2

1/2

-2/3

8.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the X component of the unit tangent vector?

4 / √13

1 / √13

2 / √13

3 / √13

9.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the Y component of the unit tangent vector?

-4 / √13

1 / √13

-2 / √13

3 / √13

10.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is it important for the X component of the tangent vector to be positive?

To ensure the vector points in the correct direction.

To make calculations easier.

To match the gradient vector.

To simplify the function.

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