Calc 3, Exam 1 walkthrough (Spring 2023)

The video begins with an introduction to a multi-variable calculus exam from Spring 2023, focusing on geometry. The instructor emphasizes the importance of writing one's name on the exam and then dives into Problem One. This problem involves finding the midpoint between two points, A and B, in a three-dimensional space. The midpoint is calculated by averaging the coordinates of the two points. The second part of the problem asks for the area of a triangle formed by three points, A, B, and C. The instructor explains that the cross product of vectors representing two sides of the triangle can be used to find the area, specifically by taking half of the magnitude of the cross product.

In this paragraph, the instructor continues the discussion on finding the area of a triangle in three-dimensional space. The method involves selecting two vectors that represent the sides of the triangle and calculating their cross product. The magnitude of this cross product, divided by two, gives the area of the triangle. The instructor demonstrates how to find these vectors by considering the changes in coordinates between the points and emphasizes the importance of careful arithmetic to avoid mistakes. The cross product is calculated using a matrix and determinant method, and the final area is obtained by taking the square root of the sum of the squares of the components of the cross product.

The instructor moves on to a problem involving a curve in three-dimensional space, described by a parametric equation. The task is to find the length of the curve from T=0 to T=8. The approach involves integrating the speed, which is the magnitude of the derivative of the position vector with respect to T. The derivative is calculated, and the magnitude is found by integrating the square root of the sum of the squares of the components of the derivative. The instructor uses a substitution method to simplify the integral and finds the length of the curve to be an integer or a fully reduced fraction.

In this section, the focus is on finding the unit tangent vector to a curve at a specific point. The unit tangent vector is derived from the derivative of the position vector, which represents the velocity. The magnitude of this derivative gives the speed, and dividing the velocity vector by its magnitude yields the unit tangent vector. The instructor demonstrates how to calculate this at T=4, resulting in a vector that points in the direction of the curve at that point.

The video script discusses the concept of vector projection, which is related to the dot product. The projection of one vector onto another is found by taking the dot product of the two vectors and dividing by the magnitude of the vector being projected onto. The instructor explains that the direction of the projection must align with the vector onto which it is being projected. The example given involves projecting a vector X onto a vector Y, resulting in a projection vector Z. The instructor also touches on the geometrical intuition behind projections and how they relate to the direction of the vectors.

The instructor addresses a problem involving an asteroid moving along a straight line in three-dimensional space. The task is to find the minimum distance from Earth to the asteroid. The asteroid's path is represented by a linear equation, and the instructor suggests several methods to solve the problem, including using the dot product and cross product. The solution involves finding a point on the asteroid's path that forms a right angle with Earth's position, which can be done by setting up an equation based on the direction of the asteroid's path and solving for the time when the distance is minimized. The final answer includes the units of parsecs.

Continuing from the previous problem, the instructor calculates the speed of the asteroid when it reaches the minimum distance from Earth. The speed is determined by taking the magnitude of the derivative of the asteroid's position vector with respect to time. Since the path is a straight line, the speed is constant, and the magnitude of the velocity vector at any point on the line is the same. The instructor concludes that the asteroid moves at a constant speed of three parsecs per year.

The script introduces a matching exercise where equations in Cartesian coordinates are paired with their corresponding equations in cylindrical or spherical coordinates. The instructor walks through the process of identifying the correct matches by understanding the relationships between the different coordinate systems. Each equation is analyzed, and the corresponding equation in the other coordinate system is identified, demonstrating a thorough understanding of the transformations between Cartesian, cylindrical, and spherical coordinates.

In this section, the instructor asks viewers to describe the graphs of various coordinate equations. The equations represent different geometric shapes such as planes, cylinders, cones, and spheres. The instructor explains how to identify the type of graph each equation represents based on the presence or absence of certain variables and the form of the equation. The descriptions provided help in understanding the geometric interpretation of the equations.

The video concludes with a problem involving a parametric curve. The task is to find the tangent line to the curve at a specific point. The instructor demonstrates how to calculate the tangent line by first identifying the point on the curve and then finding the derivative of the position vector with respect to the parameter, which gives the direction of the tangent line. The tangent line is then expressed in parametric form, using the point and the direction vector.

The final problem involves finding the equation of the line where two planes intersect. The instructor explains that the intersection of two planes is a line and provides two methods to find the equation of this line. The first method involves finding a direction vector that is perpendicular to the normal vectors of both planes using the cross product. A point on the line is then found by solving the equations of the planes simultaneously. The second method involves solving the equations for one variable in terms of another and then expressing the line in parametric form. The instructor emphasizes that there are multiple correct ways to express the equation of the line, as long as it satisfies the conditions of the planes.