Linear Models (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Water Draining from a Bucket

A bucket initially full of water has a hole in its side. The graph shows the amount of water remaining over time.

Part a: Determine the gradient of the function

To find the gradient, we use the formula:

$\text{Gradient} = \frac{\text{rise}}{\text{run}}$

From the graph, we can identify two clear points: $(0, 30)$ and $(60, 0)$.

The rise is $30$ units down, which we write as $-30$.

The run is $60$ units across.

$\text{Gradient} = \frac{-30}{60} = -\frac{1}{2}$

Part b: Determine the vertical intercept

Looking at where the line crosses the vertical axis, we can see this occurs at $y = 30$. Therefore, the vertical intercept is 30.

Part c: Explain what the gradient represents in this context

The gradient represents the rate of change between the two physical quantities being measured. In this situation, it shows how the amount of water in the bucket changes over time.

A gradient of $-\frac{1}{2}$ means the amount of water in the bucket decreases by 1 litre every 2 minutes. In other words, $1$ litre of water drains out through the hole every $2$ minutes.

Part d: Explain what the vertical intercept represents in this context

The vertical intercept represents the initial value - the starting amount before any change occurs. In this case, the vertical intercept of $30$ represents the initial amount of water in the bucket before water began draining out through the hole. The bucket initially contained 30 litres.

Part e: Determine an equation using appropriate variables

To create an equation for this situation, we should choose variables that represent the context clearly. Let's use:

• $V$ for volume (the dependent variable, measured in litres)
• $t$ for time (the independent variable, measured in minutes)

Starting with the general form $y = mx + c$, we substitute our variables:

Here, $m = -\frac{1}{2}$ (the gradient) and $c = 30$ (the vertical intercept).

The equation $V = -\frac{1}{2}t + 30$ describes the volume of water remaining in the bucket after $t$ minutes.

Part f: Determine the amount of water remaining after 20 minutes

To find this, we substitute $t = 20$ into our equation:

After $20$ minutes, there are 20 litres of water remaining in the bucket.

Worked Example: Burning Candle

A candle with height $h$ cm burns according to the equation $h = -2t + 8$, where $t$ is the time elapsed in minutes.

Part a: Complete the table of values

To complete this table, we substitute each value of $t$ into the equation:

When $t = 0$: $h = -2 \times 0 + 8 = 8$

When $t = 1$: $h = -2 \times 1 + 8 = 6$

When $t = 2$: $h = -2 \times 2 + 8 = 4$

When $t = 3$: $h = -2 \times 3 + 8 = 2$

Part b: Sketch the graph

To sketch the graph, we plot the points from our table: $(0, 8)$, $(1, 6)$, $(2, 4)$, and $(3, 2)$. Then we draw a straight line through these points.

Part c: Determine the valid domain for the model

Physical constraints affect which values make sense in this model. Both the height of a candle and elapsed time cannot be negative values in reality. Therefore, we need $h \geq 0$ and $t \geq 0$.

Looking at the graph in the first quadrant, we can see the model is only valid when 0 ≤ t ≤ 4. When $t = 4$, the equation gives $h = -2(4) + 8 = 0$, meaning the candle has completely burned down. The equation accurately models the candle's height in the first quadrant only.

Worked Example: Carpenter's Charges

A carpenter charges a call-out fee of $150 plus $45 per hour for work.

Part a: Determine an equation for the total amount charged

We need to express the relationship:

Total amount charged = $45 × Number of hours + $150

Using appropriate variables:

• Let $C$ represent the total cost (dependent variable)
• Let $h$ represent the number of hours worked (independent variable)

Starting with the general form and substituting:

Here, $m = 45$ (the gradient, representing the hourly rate) and $c = 150$ (the vertical intercept, representing the call-out fee).

Part b: Determine the total amount charged for 6 hours of work

We substitute $h = 6$ into the equation:

The total amount charged is $420.

Part c: Determine the number of hours worked if the total charge is $735

This time we know the cost and need to find the hours. We substitute $C = 735$ and solve for $h$:

The carpenter worked for 13 hours.

Worked Example: Hairstyling Cost

Calandra charges $54.00 to style hair, plus an additional $8.50 per foil. Pauline wants the total cost to be no more than $137.00.

Part a: Write an inequality representing the number of foils Pauline could get

The phrase "no more than" means "less than or equal to." We can express this situation in words:

Cost of styling + Cost per foil × Number of foils ≤ Total Pauline can spend

Let $N$ represent the number of foils. Translating into an algebraic inequality:

$54.00 + 8.50N \leq 137.00$

Part b: How many foils could Pauline get and still afford the styling?

We solve the inequality:

According to the mathematical solution, Pauline could get $9.76$ foils or fewer. However, since she cannot get a partial foil, a more realistic solution is that she can get 9 foils or fewer.

Part c: Determine whether $N = -2$ is a viable solution

While $N = -2$ is part of the mathematical solution set for the inequality $N \leq 9.76$, it is not a viable solution in this context. It doesn't make sense to have a negative number of foils. Pauline can get a maximum of $9$ foils and a minimum of $0$ foils, so $-2$ is not realistic.

This demonstrates an important point: a non-viable solution satisfies the inequality mathematically but is not possible in the real-world context. Always interpret your mathematical solutions within the context of the problem.