Summary
Highlights
The video introduces combined variation, building upon previous lessons on direct, inverse, and joint variations. Combined variation involves both direct and inverse relationships between variables, and sometimes joint variation as well.
A quick review of direct variation (y = kx), inverse variation (y = k/x), and joint variation (a = kbc) is provided to set the foundation for understanding combined variation, where 'k' is the constant of variation.
Combined variation is defined as a relationship among variables that involves both direct and inverse variation, and sometimes joint variation. An example provided is 'z varies directly as x and inversely as y', which translates to z = kx/y.
The first example demonstrates translating the statement 'w varies jointly as c and the square of a and inversely as b' into the equation w = k(ca²)/b. This highlights how to incorporate joint and inverse variation in one equation.
The second example translates 'p varies directly as the square of x and inversely as s' into the equation p = kx²/s. This reinforces the understanding of direct and inverse relationships.
This example relates to physics: 'The electrical resistance R of a wire varies directly as the length L and inversely as the square of its diameter D'. This translates to R = kL/D², showing application in a real-world context.
Another physics application: 'The acceleration A of a moving object varies directly as the distance D it travels and inversely as the square of the time T it travels'. The equation becomes A = kD/T².
The final translation example covers 'The pressure P of a gas varies directly at its temperature T and inversely at its volume V', resulting in the equation P = kT/V.
A problem is presented: 'If z varies directly as x and inversely as y, and z = 9 when x = 6 and y = 2, find z when x = 8 and y = 12.' The first step is to translate the statement to z = kx/y and then use the given values to solve for 'k', which is found to be 3.
After finding the constant of variation (k=3), the video proceeds to find the value of 'z' using the new given values (x=8, y=12) and the derived equation z = 3x/y. The calculation shows z = 2.
A more complex problem: 'If R varies directly as S and inversely as the square of U, and R = 2 when S = 18 and U = 2, find R when U = 3 and S = 27.' The equation is R = kS/U². Substituting the initial values, 'k' is found to be 4/9 after simplification.
Using the constant k = 4/9 and the equation R = (4/9)S/U², the value of 'R' is calculated when S = 27 and U = 3. Substituting these values, R is found to be 4/3.
Given k = 4/9 and the equation R = (4/9)S/U², the video demonstrates how to find 'S' when R = 4 and U = 2. After substitution and calculation, 'S' is found to be 36.
The final example shows how to find 'U' when R = 1 and S = 36, using the same constant of variation and equation. After substituting and solving, 'U' is found to be 4.