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Difficulty: Very hardSlope of a Line

A system administrator monitors the temperatures of two servers, Server A and Server B, over a period of tt hours after midnight. The temperature of Server A, in degrees Celsius, is modeled by a linear function of time. At t=2t = 2 hours, the temperature is 30∘C30^\circ\text{C}, and at t=at = a hours (where a>2a > 2), the temperature is (30+a2−2a)∘C(30 + a^2 - 2a)^\circ\text{C}. The temperature of Server B, in degrees Celsius, is also modeled by a linear function of time. At t=1t = 1 hour, Server B's temperature is 25∘C25^\circ\text{C}, and at t=at = a hours, its temperature is (a2+30)∘C(a^2 + 30)^\circ\text{C}. If the temperature of Server B increases at a constant rate that is 1.51.5 times the constant rate of temperature increase of Server A, what is the value of aa?

  1. A
    2
  2. B
    3
  3. 5Answer
  4. D
    11
  5. E
    13

Answer

The value of aa is 5.
To find the rate of change (slope) for each server, we apply the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. For Server A, the slope is mA=(30+a2−2a)−30a−2=am_A = \frac{(30 + a^2 - 2a) - 30}{a - 2} = a. For Server B, the slope is mB=(a2+30)−25a−1=a2+5a−1m_B = \frac{(a^2 + 30) - 25}{a - 1} = \frac{a^2 + 5}{a - 1}. We are given that mB=1.5mAm_B = 1.5 m_A, which yields the equation a2+5a−1=1.5a\frac{a^2 + 5}{a - 1} = 1.5a. Multiplying both sides by a−1a - 1 gives a2+5=1.5a(a−1)=1.5a2−1.5aa^2 + 5 = 1.5a(a - 1) = 1.5a^2 - 1.5a. Grouping all terms on one side gives 0.5a2−1.5a−5=00.5a^2 - 1.5a - 5 = 0. Multiplying the equation by 2 results in a2−3a−10=0a^2 - 3a - 10 = 0, which factors as (a−5)(a+2)=0(a - 5)(a + 2) = 0. Since a>2a > 2, the only valid solution is the positive root, which is 5.

Step-by-Step Solution

1
Calculate the constant rate of temperature increase (slope) for Server A, mAm_A, using the coordinates (2,30)(2, 30) and (a,30+a2−2a)(a, 30 + a^2 - 2a).
mA=(30+a2−2a)−30a−2=a2−2aa−2=a(a−2)a−2=am_A = \frac{(30 + a^2 - 2a) - 30}{a - 2} = \frac{a^2 - 2a}{a - 2} = \frac{a(a-2)}{a-2} = a (since a>2a > 2)
The constant rate of change of a linear temperature model is represented by the slope of the line passing through its data points.
2
Calculate the constant rate of temperature increase (slope) for Server B, mBm_B, using the coordinates (1,25)(1, 25) and (a,a2+30)(a, a^2 + 30).
mB=(a2+30)−25a−1=a2+5a−1m_B = \frac{(a^2 + 30) - 25}{a - 1} = \frac{a^2 + 5}{a - 1}
The constant rate of change of Server B is represented by the slope of the line passing through its data points.
3
Set up the equation using the given relationship that Server B's rate of increase is 1.51.5 times Server A's rate (mB=1.5mAm_B = 1.5 m_A).
1.5a=a2+5a−11.5a = \frac{a^2 + 5}{a - 1}
Equating the rate of change of Server B to 1.5 times the rate of change of Server A allows us to write an equation in terms of the variable aa.
4
Solve the equation for aa by cross-multiplying and simplifying the resulting quadratic equation.
1.5a(a−1)=a2+5  ⟹  1.5a2−1.5a=a2+5  ⟹  0.5a2−1.5a−5=0  ⟹  a2−3a−10=0  ⟹  (a−5)(a+2)=01.5a(a - 1) = a^2 + 5 \implies 1.5a^2 - 1.5a = a^2 + 5 \implies 0.5a^2 - 1.5a - 5 = 0 \implies a^2 - 3a - 10 = 0 \implies (a-5)(a+2) = 0
Cross-multiplying eliminates the fraction, and grouping all terms on one side forms a standard quadratic equation that can be factored.
5
Identify the valid value of aa that satisfies the initial domain constraint.
a=5a = 5 (since a=−2a = -2 is rejected because the problem states a>2a > 2)
The solution must satisfy the given condition that a>2a > 2 for the time intervals to be valid.

Key Concept

Calculating and equating rates of change (slopes) of linear functions using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} and solving the resulting quadratic equation.
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Topics

Slope of a Line

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