Summary
Highlights
The video begins by introducing joint variation, defining it as 'a varies jointly as b and c' meaning a = kbc, where k is the constant of variation. Objectives include defining, representing, solving for constants and unknowns, and solving problems in joint variation.
Several examples are provided to illustrate joint variation, such as the area of a triangle varying jointly as its base and altitude (A = kbh, where k = 1/2) and the pressure of a gas varying jointly as its density and absolute temperature (P = kDT).
The video demonstrates how to translate verbal statements into mathematical equations using 'k' as the constant of variation. Examples include 'P varies jointly as Q and R' becoming P = kQR, and 'the volume of a cylinder V varies jointly as its height h and the square of the radius r' becoming V = khr².
This section focuses on solving for the constant of variation and then writing the complete equation. For example, if A varies jointly as B and C, and A=36 when B=3 and C=4, the constant k is found to be 3, leading to the equation A = 3BC.
The video provides an example of finding the area of a triangle given new base and altitude values after determining the constant of variation from initial conditions. This involves calculating 'k' (e.g., 1/2 for the area of a triangle) and then substituting it along with new values to find the unknown.
Combined variation is introduced as a combination of direct and inverse variation. The definition is 'Z varies directly as X and inversely as Y' meaning Z = kX/Y, where k is the constant of variation. This type of variation involves more than two variables.
Examples are given for translating combined variation statements into mathematical sentences, such as 'T varies directly as A and inversely as B' (T = kA/B) and 'Y varies directly as X and inversely as the square of Z' (Y = kX/Z²).
Practical problems involving combined variation are solved. For instance, if Z varies directly as X and inversely as Y, and Z=9 when X=6 and Y=2, the constant k is found (k=3). Then, Z is calculated for new values of X and Y (X=8, Y=12), resulting in Z=2.
Another problem illustrates finding a value when the variation involves a square. If T varies directly as M and inversely as the square of N, and T=16 when M=8 and N=2, the constant k is determined to be 8. Then, T is calculated for new values (M=13, N=3), yielding T=104/9.