Problem 50 - Algebra Form and Function

Question

A boy's height \(h,\) in feet, \(t\) years after his \(10^{\text {th }}\) birthday, is given by \(h=4+0.2 t\). Which of the following equivalent expressions for this function shows most clearly his height at age 20 ? What is that height? (i) \(\quad h=4+0.2 t\) (ii) \(\quad h=6+0.2(t-10)\) (iii) \(\quad h=10+0.2(t-30)\)

Step-by-Step Solution

Answer

Answer: The expression most clearly showing the boy's height at age 20 is Expression (ii): \(h=6+0.2(t-10)\). The boy's height at age 20 is 6 feet.

Step 1: Understand each expression

Each of the three expressions are equivalent ways to represent the boy's height. They all represent the height function in terms of t, where t is the years after his 10th birthday. (i) The first expression is given by \(h=4+0.2t\). (ii) The second expression is given by \(h=6+0.2(t-10)\). (iii) The third expression is given by \(h=10+0.2(t-30)\).

Step 2: Find the boy's height at age 20

To find the boy's height at age 20, we need to set t = 10 (since he is 20 years old and the function is given in terms of years after his 10th birthday). We will plug in t = 10 into each expression and see which one simplifies the calculation most clearly. (i) \(h=4+0.2(10)=4+2=6\) feet (ii) \(h=6+0.2(10-10)=6+0.2(0)=6\) feet (iii) \(h=10+0.2(10-30)=10+0.2(-20)=10-4=6\) feet

Step 3: Determine the expression that shows the height most clearly at age 20

We found out that the boy's height is 6 feet at age 20 for each expression. However, Expression (ii) shows the height most clearly when t = 20, as we directly get 6 feet without any further calculations.

Step 4: Conclusion

Therefore, the expression most clearly showing the boy's height at age 20 is Expression (ii): \(h=6+0.2(t-10)\) The boy's height at age 20 is 6 feet.

Key Concepts

Equivalent Expressions

Equivalent expressions are different ways to write the same mathematical relationship or equation, using the same variables. They hold the same value for all values of the variables involved. This concept is useful because sometimes one form is better suited for specific calculations or interpretations.
In the context of the boy's height problem:

• Expression (i): \(h = 4 + 0.2t\) is straightforward and shows height based on years after his 10th birthday.
• Expression (ii): \(h = 6 + 0.2(t-10)\) adjusts the base value, highlighting the height more clearly when the boy reaches age 20.
• Expression (iii): \(h = 10 + 0.2(t-30)\) restructures the terms in a way some might find less intuitive without context.

Although these expressions might look different, they represent the same height function. Recognizing equivalent expressions allows us to choose the form that simplifies the task at hand.

Function Notation

Function notation is a way to express relations between input and output using specific symbols and letters. The height function, in this case, is described using the letter \(h\) to represent the output (height), while \(t\) represents the input (years since the 10th birthday). Function notation is written as \(h = f(t)\), where \(f\) indicates the functional relationship between \(h\) and \(t\).
This method of notation helps make functions clear and consistent:

• \(h = 4 + 0.2t\) directly ties each year to an increase in height.
• The same goes for other forms, like \(h = 6 + 0.2(t-10)\), which shows the adjusted starting point.

Essentially, function notation helps us understand how changes in the input (like increasing age) affect the output (height) in a reliable manner.

Solving Equations

Solving equations involves finding the values of variables that make an equation true. In the height problem, we solve equations to find the boy's height at age 20 by setting \(t=10\) because the equation expresses height in years after his 10th birthday.
Let's look at how to solve each expression:

• For \(h = 4 + 0.2t\), substitute \(t = 10\): \(h = 4 + 0.2 \times 10 = 6\) feet.
• For \(h = 6 + 0.2(t-10)\), substitute \(t = 10\): \(h = 6 + 0.2 \times (10-10) = 6\) feet, done more straightforwardly.
• For \(h = 10 + 0.2(t-30)\), substitute \(t = 10\): \(h = 10 + 0.2 \times (10-30) = 6\) feet.

Through solving these equations, students can see how to manipulate expressions to achieve the desired outcome, highlighting which form simplifies the process the most.