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Direct Inverse and Joint Variation Word Problems

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Summary

This video covers how to write and solve direct, inverse, and joint variation word problems. It explains how to set up the equations, solve for the constant of proportionality (K), and then use that constant to find unknown values in different scenarios. The video demonstrates both conceptual and algebraic approaches to solving these problems, including variations involving squares, cubes, and square roots.

Highlights

Solving an Inverse Variation Problem
00:05:08

This section tackles an inverse variation problem: 'Y varies inversely with X. If X is 4 when Y is 48, what is Y when X is 8?' The equation is Y=K/X. Solving for K gives K=192. Using K, the new Y value is found to be 24. Conceptually, for inverse variation, if X doubles, Y halves.

Inverse Variation with a Cube
00:13:14

The video presents an inverse variation problem involving a cube: 'Y varies inversely with the cube of X. If Y is 108 when X is 2, what is Y when X is 6?' The equation is Y=K/X^3. K is calculated as 864. Using K and the new X, Y is found to be 4. Conceptually, if X triples, Y decreases by a factor of X^3, which is 3^3=27.

Solving a Joint Variation Problem
00:08:03

The video then moves to a joint variation problem: 'Y varies jointly with X and Z. If Y is 36 when X is 2 and Z is 3, what is Y when X is 4 and Z is 6?' The equation is Y=KXZ. First, K is calculated as 6. Then, using the new values, Y is found to be 144. The conceptual explanation highlights that if X doubles and Z doubles, Y increases by a factor of 2x2=4.

Direct Variation with a Square
00:11:09

This part illustrates a direct variation problem with a squared term: 'Y varies directly with the square of X. If Y is 8 when X is 2, what is Y when X is 4?' The equation is Y=Kx^2. Finding K yields K=2. Substituting the new X value, Y is determined to be 32. Conceptually, if X doubles, Y increases by a factor of X^2, which is 2^2=4.

Introduction to Variation Equations
00:00:00

The video starts by explaining how to write equations for direct, inverse, and joint variation. Direct variation is represented as Y = KX, inverse variation as Y = K/X, and joint variation as Y = KXZ. It also covers more complex variations like direct with a square (Y = Kx^2), inverse with a square root (Y = K/sqrt(R)), and combined variations where one variable is direct and another is inverse.

Solving a Direct Variation Problem
00:03:07

The first example demonstrates a direct variation problem: 'Y varies directly with X. If X is 3 when Y is 12, what is Y when X is 9?' The process involves writing the equation (Y=KX), solving for K using the given values (K=4), and then using K to find the new Y value (Y=36). The video also explains the conceptual understanding of direct proportionality, where if X triples, Y also triples.

Combined Direct and Inverse Variation
00:16:41

The final problem combines direct and inverse variation: 'Y varies directly with the cube of X and inversely with the square of Z. If Y is 8 when X is 2 and Z is 3, find Y when X is 4 and Z is 6.' The equation is Y = Kx^3/Z^2. K is found to be 9. With the new X and Z values, Y is calculated as 16. The conceptual approach involves multiplying the direct effect (X cubed) and dividing by the inverse effect (Z squared).

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