Problem 74 - Algebra and Trigonometry

Question

Changing Temperature Scales The temperature on a certain afternoon is modeled by the function $$C(t)=\frac{1}{2} t^{2}+2$$ where \(t\) represents hours after 12 noon \((0 \leq t \leq 6),\) and \(C\) is measured in "C. (a) What shifting and shrinking operations must be performed on the function \(y=t^{2}\) to obtain the function \(y=C(t) ?\) (b) Suppose you want to measure the temperature in \(^{\circ} \mathrm{F}\) instead. What transformation would you have to apply to the function \(y=C(t)\) to accomplish this? (Use the fact that the relationship between Celsius and Fahrenheit degrees is given by \(F=\frac{9}{5} C+32 .\) ) Write the new function \(y=F(t)\) that results from this transformation.

Step-by-Step Solution

Answer

(a) Vertical shrink by 1/2 and shift up 2 units. (b) \(F(t) = \frac{9}{10} t^2 + \frac{178}{5}\).

Step 1: Analyze Shifting and Shrinking Operations

To find the shifting and shrinking operations, we compare the given function \(C(t) = \frac{1}{2} t^2 + 2\) with \(y = t^2\):1. The term \(\frac{1}{2} t^2\) suggests a vertical shrink by a factor of \(\frac{1}{2}\) compared with \(y = t^2\).2. The number \(+2\) indicates the entire function shifts upwards by 2 units.Thus, we have a vertical shrink by \(\frac{1}{2}\) and an upward shift by 2 units.

Step 2: Convert Celsius to Fahrenheit

To convert the temperature from Celsius to Fahrenheit, we use the conversion formula:\[F = \frac{9}{5}C + 32\]We substitute \(C(t) = \frac{1}{2} t^2 + 2\) into this equation:\[F(t) = \frac{9}{5}\left(\frac{1}{2} t^2 + 2\right) + 32\]

Step 3: Simplify the Fahrenheit Function

Simplifying the equation obtained in the last step:1. Distribute the \(\frac{9}{5}\): \[F(t) = \frac{9}{5} \times \frac{1}{2} t^2 + \frac{9}{5} \times 2 + 32\] \[= \frac{9}{10} t^2 + \frac{18}{5} + 32\]2. Combine the constant terms: First, convert \(32\) to fifths: \(32 = \frac{160}{5}\) \[\frac{18}{5} + \frac{160}{5} = \frac{178}{5}\]Thus, the new function is:\[F(t) = \frac{9}{10} t^2 + \frac{178}{5}\]

Key Concepts

Shifting Operations in Functions

Shifting operations in mathematical functions are useful for moving a graph up, down, left, or right. In our exercise, we start with the basic function of a parabola, which is represented by \( y = t^2 \). When we compare this with our given function \( C(t) = \frac{1}{2} t^2 + 2 \), we can determine the changes that were made.

• The term \(+2\) at the end of \( C(t) \) indicates that the entire function moves 2 units upward on the graph.
• This upward movement is called a vertical shift, and it doesn't affect the shape of the graph, just its position.

By understanding function shifts, students can visualize how graphs transform and move around the coordinate plane. This concept is foundational in understanding more complex transformations in mathematics.

Shrinking Operations in Functions

Shrinking, or compressing, operations in functions adjust the size of the graph. This is achieved by multiplying the function by a constant. In our function \( C(t) = \frac{1}{2} t^2 + 2 \), the term \( \frac{1}{2} \) in front of \( t^2 \) signifies this operation.

• This multiplier compresses the parabola vertically by a factor of \( \frac{1}{2} \).
• Vertical shrinking makes the parabola appear wider than the original \( y = t^2 \).

Comprehending shrinking transformations helps students understand how multipliers affect the look and scale of a graph. The resulting alteration maintains the parabola's shape but changes its steepness along the y-axis.

Celsius to Fahrenheit Conversion

Converting between Celsius and Fahrenheit is a common mathematical problem involving linear transformations. The relationship is defined by the formula \( F = \frac{9}{5} C + 32 \), where \( F \) is Fahrenheit and \( C \) is Celsius.

In our exercise, we start with \( C(t) = \frac{1}{2} t^2 + 2 \) and we need to convert it to \( F(t) \).

• We begin by applying the conversion formula, substituting \( C(t) \) into \( F = \frac{9}{5} C + 32 \).
• This results in the new function: \[ F(t) = \frac{9}{5} \left( \frac{1}{2} t^2 + 2 \right) + 32 \]
• The Fahrenheit version simplifies further to: \[ F(t) = \frac{9}{10} t^2 + \frac{178}{5} \]

Transforming temperature scales involves linear functions and highlights the importance of constants and coefficients in altering outputs. Mastering these conversions ensures students can navigate real-world scenarios involving temperature.