krit.club logo

Number - Constant of proportionality and graphs of direct and inverse proportion-extended

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

In direct proportion, the relationship between two variables yy and xx is defined by y=kxy = kx. Graphically, this is represented by a straight line passing through the origin (0,0)(0,0). The constant of proportionality kk represents the gradient of the line.

A linear graph passing through the origin representing direct proportion.
•

In inverse proportion, the relationship is defined by y=kxy = \frac{k}{x} or xy=kxy = k. As one variable increases, the other decreases. The graph is a hyperbola that never touches the axes (asymptotic to x=0x=0 and y=0y=0).

A hyperbolic curve representing inverse proportion.
•

Direct proportion to a square, y=kx2y = kx^2, results in a parabolic curve starting at the origin. The rate of change in yy increases as xx increases.

A parabolic graph representing direct proportion to the square.
•

Identifying kk from a graph: For a direct proportion graph, pick any point (x,y)(x, y) on the line (except the origin) and calculate k=yxk = \frac{y}{x}. For an inverse proportion graph, pick any point and calculate k=xyk = xy.

📐Formulae

y=kxy = kx

y=kxy = \frac{k}{x}

k=yx (Direct)k = \frac{y}{x} \text{ (Direct)}

k=xy (Inverse)k = xy \text{ (Inverse)}

y=kxn (Direct to a power)y = kx^n \text{ (Direct to a power)}

y=kxn (Inverse to a power)y = \frac{k}{x^n} \text{ (Inverse to a power)}

💡Examples

Problem 1:

The variable yy is directly proportional to the square of xx. When x=3x = 3, y=18y = 18. Find the value of yy when x=5x = 5.

Solution:

  1. Set up the equation: y=kx2y = kx^2
  2. Substitute x=3x = 3 and y=18y = 18: 18=k(32)18 = k(3^2) 18=9k18 = 9k k=189=2k = \frac{18}{9} = 2
  3. Write the specific equation: y=2x2y = 2x^2
  4. Substitute x=5x = 5: y=2(52)y = 2(5^2) y=2(25)=50y = 2(25) = 50

Explanation:

Since yy is proportional to x2x^2, we find the constant kk first using the given values, then use that constant to calculate the new value of yy.

Problem 2:

The pressure PP of a gas is inversely proportional to its volume VV. When V=4V = 4 m3m^3, P=100P = 100 PaPa. Find the pressure when the volume is 88 m3m^3.

Solution:

  1. Set up the equation: P=kVP = \frac{k}{V}
  2. Substitute V=4V = 4 and P=100P = 100: 100=k4100 = \frac{k}{4} k=100×4=400k = 100 \times 4 = 400
  3. Write the specific equation: P=400VP = \frac{400}{V}
  4. Substitute V=8V = 8: P=4008=50P = \frac{400}{8} = 50

Explanation:

In inverse proportion, the product of the variables is constant (P×V=kP \times V = k). Increasing the volume results in a proportional decrease in pressure.

Problem 3:

A graph of yy against xx is a straight line through the origin. If the line passes through the point (4,12)(4, 12), determine the constant of proportionality and the equation of the line.

Solution:

  1. Since the graph is a straight line through the origin, y=kxy = kx.
  2. Substitute the coordinates (4,12)(4, 12): 12=k(4)12 = k(4) k=124=3k = \frac{12}{4} = 3
  3. The equation is y=3xy = 3x.

Explanation:

The gradient of a direct proportion graph is the constant of proportionality kk. Use the point (x,y)(x, y) to find the slope.

Problem 4:

The graph shows the relationship between variables yy and xx where yy is directly proportional to x3x^3. Given that the graph passes through (2,40)(2, 40), find the value of kk and calculate yy when x=3x = 3.

A cubic curve passing through the point (2, 40).

Solution:

  1. Set up the equation: y=kx3y = kx^3.
  2. Substitute the point (2,40)(2, 40): 40=k(2)340 = k(2)^3.
  3. Solve for kk: 40=8k  ⟹  k=540 = 8k \implies k = 5.
  4. Find yy when x=3x = 3: y=5(3)3=5×27=135y = 5(3)^3 = 5 \times 27 = 135.

Explanation:

Since yy is proportional to x3x^3, we use the cubic model. Substituting the known coordinates allows us to find the constant kk, which remains the same for all points on this curve.

Problem 5:

The intensity of light II is inversely proportional to the square of the distance dd from the source. At a distance of 33 meters, the intensity is 2020 units. Find the intensity at a distance of 55 meters.

An inverse square curve showing intensity decreasing as distance increases.

Solution:

  1. State the relationship: I=kd2I = \frac{k}{d^2}.
  2. Find kk: 20=k32  ⟹  20=k9  ⟹  k=18020 = \frac{k}{3^2} \implies 20 = \frac{k}{9} \implies k = 180.
  3. Formulate the equation: I=180d2I = \frac{180}{d^2}.
  4. Calculate for d=5d = 5: I=18052=18025=7.2I = \frac{180}{5^2} = \frac{180}{25} = 7.2 units.

Explanation:

This follows the inverse square law. We first determine the constant of proportionality by substituting the initial conditions, then use the constant to evaluate the second state.