Problem 90 - 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 \(^{\circ} \mathrm{F}\) in- stead. 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

Shrink \(t^{2}\) by \(\frac{1}{2}\) and shift up by 2; use \(F(t) = \frac{9}{10}t^{2} + 35.6\).

Step 1: Identify Transformations to Obtain y=C(t)

The function given is \(C(t) = \frac{1}{2} t^{2} + 2\). To transform \(y = t^{2}\) into \(y = C(t)\), we need to apply two operations:- Shrink \(t^{2}\) vertically by a factor of \(\frac{1}{2}\), changing \(t^{2}\) to \(\frac{1}{2}t^{2}\).- Shift the graph upward by 2 units to account for the "+2" term in \(C(t)\).

Step 2: Convert Celsius to Fahrenheit

To convert the Celsius function \(C(t)\) to Fahrenheit, we use the relationship: \[ F = \frac{9}{5}C + 32 \]Substitute \(C(t)\) into this equation to transform \(C(t)\) into \(F(t)\):\[ F(t) = \frac{9}{5}\left(\frac{1}{2}t^{2} + 2\right) + 32 \]

Step 3: Simplify the Fahrenheit Function

Further simplify the expression obtained in Step 2:\[F(t) = \frac{9}{5}\left(\frac{1}{2}t^{2} + 2\right) + 32= \frac{9}{5} \cdot \frac{1}{2} t^{2} + \frac{9}{5} \cdot 2 + 32= \frac{9}{10} t^{2} + \frac{18}{5} + 32= \frac{9}{10} t^{2} + 3.6 + 32= \frac{9}{10} t^{2} + 35.6\]Thus, the new function \(F(t)\) is \(F(t) = \frac{9}{10} t^{2} + 35.6\).

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial function where the highest exponent of the variable is two. These functions are generally expressed in the form \( y = ax^2 + bx + c \), where \( a eq 0 \). The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the sign of \( a \). Here are some important characteristics of quadratic functions:

• The vertex, which is the highest or lowest point of the parabola.
• The axis of symmetry, a vertical line that passes through the vertex and divides the parabola into two mirror-image halves.
• The roots or zeros, which are the points where the parabola intersects the x-axis, if any exist.

For example, consider the function \( y = t^2 \), which is a simple parabola opening upwards with its vertex at the origin (0,0). By altering this basic function with various transformations, different specific quadratic functions can be formed. Understanding these fundamental properties of quadratic functions is crucial for performing any transformations or solving practical problems that involve such functions.

Function Transformations

Function transformations involve changing the basic properties of a function to modify its graph. In the case of the given function \( y = C(t) = \frac{1}{2}t^2 + 2 \), we start with the basic quadratic function \( y = t^2 \). We need to apply specific transformations to morph it into our desired form. The two transformations applied are:

• Vertical Shrink: Multiply \( t^2 \) by \( \frac{1}{2} \). This operation compresses the graph vertically, making it narrower. The vertical stretch or shrink is governed by the coefficient of \( t^2 \), in this case, \( \frac{1}{2} \).
• Vertical Shift: Add 2 to the function. This moves the entire graph upwards by 2 units. Vertical shifts are controlled by adding or subtracting a constant term from the function.

This process of transforming the function allows us to model different scenarios easily by tweaking the mathematical expressions. In practice, such transformations help in aligning the modelled function more closely with real-world data or required specifications.

Celsius to Fahrenheit

Converting temperatures between Celsius and Fahrenheit is a common task and involves a specific linear transformation. The relationship between the two measurement scales is given by the formula: \[ F = \frac{9}{5}C + 32 \] This conversion can be understood as:

• Scaling Factor: The term \( \frac{9}{5} \) is a scaling factor that adjusts the unit size, reflecting the difference in the interval divisions of Celsius and Fahrenheit scales.
• Offset: Adding 32 adjusts for the different starting points (zero points) of the two scales.

To convert the function \( C(t) \) to \( F(t) \), we substitute it into the conversion formula, which transforms it into a quadratic function in terms of Fahrenheit:\[ F(t) = \frac{9}{10} t^2 + 35.6 \] This new function represents the temperature in degrees Fahrenheit over time, again with a quadratic shape modified for unit conversion. This formula is particularly handy for applications that require frequent switching between temperature units, such as weather modeling or cooking instructions.