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Chapter 12: Problem 38

A production formula for a student's performance on a difficult English examination is given by $$g=4 h x-0.2 h^{2}-10 x^{2}$$ where \(g\) is the grade the student can expect to obtain, \(h\) is the number of hours of study for the examination, and \(x\) is the student's grade point average. The instructor finds that students' grade point averages have remained constant at \(3.0\) over the years, and that students currently spend an average of 15 hours studying for the examination. However, scores on the examination are dropping at a rate of 10 points per year. At what rate is the average study time decreasing?

Short Answer

Expert verified
The average study time is decreasing at a rate of \(-2\) hours per year, causing students' exam scores to decrease by 10 points per year.

Step by step solution

01

List the known variables and constants

We are given the following information: - Grade point average (x) = 3.0 (constant) - Average study time (h) is decreasing - Rate of decrease in examination score (dg/dt) = -10 points per year - g is given by the formula: \(g = 4hx - 0.2h^2 - 10x^2\) #Step 2: Implicit Differentiation#
02

Differentiate the production formula with respect to time

Since we are trying to find the rate at which the average study time (h) is decreasing, we need to differentiate the production formula with respect to time (t). Using implicit differentiation, we get: $$\frac{dg}{dt} = \frac{\partial g}{\partial h} \frac{dh}{dt} + \frac{\partial g}{\partial x} \frac{dx}{dt}$$ Since x is constant, its derivative with respect to time is zero. Now, we differentiate the production formula with respect to h: $$\frac{\partial g}{\partial h} = 4x - 0.4h$$ Substitute the known values of x and dg/dt: $$-10 = (4 \cdot 3 - 0.4h) \frac{dh}{dt}$$ #Step 3: Solve for the rate of change of average study time#
03

Solve for dh/dt

Now, we need to find \(\frac{dh}{dt}\), which represents the rate at which the average study time is decreasing. Simplify the equation: $$-10 = (12 - 0.4h) \frac{dh}{dt}$$ We are given that students currently spend an average of 15 hours studying for the examination. Now plug in the value of h: $$-10 = (12 - 0.4 \cdot 15) \frac{dh}{dt}$$ Solve for \(\frac{dh}{dt}\): $$\frac{dh}{dt} = \frac{-10}{12 - 6}$$ $$\frac{dh}{dt} = -2$$ #Step 4: Interpret the result#
04

Explain the meaning of the result

The rate at which the average study time is decreasing is -2 hours per year. This means that every year, students on average are spending 2 hours less studying for the examination, which is causing their scores to decrease by 10 points per year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Implicit Differentiation
Implicit differentiation is a technique used in calculus when dealing with equations where the dependent and independent variables are not easily separable. In simpler terms, it helps us find the rate of change for one variable concerning another, even when those variables are tangled together in an equation.

In the context of this exercise, we have a production formula that mixes variables for grades (\(g\)), study hours (\(h\)), and grade point average (\(x\)). To determine how the rate of study hours (\(h\)) changes over time, we utilize implicit differentiation. This approach allows us to differentiate the given function with respect to time (\(t\)), even when time (\(t\)) doesn't directly appear in the formula.

Here's a quick rundown of the steps involved:
  • Consider each of the variables as a function of time.
  • Differentiate both sides of the equation concerning time (\(t\)).
  • You apply the chain rule to variables that depend on time.
Ultimately, solving the differentiated equation gives us the rate of change of study hours. This technique is incredibly handy when encountering similar real-world problems and modeling scenarios.
Rate of Change
The rate of change is a crucial concept in calculus that helps describe how one quantity changes concerning another. In this scenario, it represents how the students' average study time (\(h\)) changes as the year progresses. Understanding the rate of change assists us in making predictions and drawing conclusions about trends.

In our problem, the given rate of change for the exam score (\(\frac{dg}{dt}\)) is a decrease of 10 points per year. By understanding this information and applying it to the production formula, we can deduce how the study time (\(h\))鈥攁nother rate of change鈥攙aries accordingly. After differentiating and solving, we find out that (\(\frac{dh}{dt}\)) or the rate of decrease of study time is 2 hours per year.

This piece of information doesn't mean much alone, but considering scores are dropping by 10 points each year, we ascertain that less study time could likely be a significant factor. Identifying these rates of change highlights areas of concern and gives insight into possible solutions, like encouraging more study hours to counteract the score drop.
Mathematical Modeling
Mathematical modeling is about crafting real-world problems into mathematical forms so they can be analyzed and solved. In this scenario, variables such as study time and grade averages are molded into a formula to predict expected grades. This model is advantageous because it combines different factors affecting an outcome and examines their interplay.

Our formula here is a simple model that provides insights into how grades are influenced by study hours and grade point averages. This particular model pulls us into the calculus realm, where implicit differentiation and rates of change come into play. Modeling serves as a bridge between abstract mathematical concepts and tangible, everyday situations.

In educational settings, such models can assist in predicting outcomes, planning strategies, or identifying discrepancies like the decline in scores here. When changes occur, like decreasing study hours, the model can highlight this variable's influence on overall scores, providing a basis for interventions.

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Most popular questions from this chapter

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