Math Problem Statement

The fuel efficiency in miles per gallon for vehicles in the USA can be modelled by the function E(t) = -0.0007t^3 + 0.0278t^2 - 0.0843t + 12 where t is years since 1970, 0 less than or equal to t less than or equal to 35. (a) Find E'(t) (b) Fuel efficiency was improving (increasing) between the years a and b. Find the values of a and b.

Solution

(a) Find E(t)E'(t)

The given function for fuel efficiency is:

E(t)=0.0007t3+0.0278t20.0843t+12E(t) = -0.0007t^3 + 0.0278t^2 - 0.0843t + 12

To find E(t)E'(t), we take the derivative of each term:

  1. The derivative of 0.0007t3-0.0007t^3 is 0.00073t2=0.0021t2-0.0007 \cdot 3t^2 = -0.0021t^2
  2. The derivative of 0.0278t20.0278t^2 is 0.02782t=0.0556t0.0278 \cdot 2t = 0.0556t
  3. The derivative of 0.0843t-0.0843t is 0.0843-0.0843
  4. The derivative of the constant 1212 is 00

Thus, the derivative E(t)E'(t) is:

E(t)=0.0021t2+0.0556t0.0843E'(t) = -0.0021t^2 + 0.0556t - 0.0843


(b) Find the values of aa and bb

Fuel efficiency is increasing when E(t)>0E'(t) > 0. To find when this occurs, we need to solve E(t)>0E'(t) > 0.

The equation is:

0.0021t2+0.0556t0.0843>0-0.0021t^2 + 0.0556t - 0.0843 > 0

Let's solve this quadratic inequality.

First, solve the corresponding quadratic equation 0.0021t2+0.0556t0.0843=0-0.0021t^2 + 0.0556t - 0.0843 = 0 to find the critical points.The critical points are approximately t=1.61t = 1.61 and t=24.86t = 24.86.

Now, to determine when fuel efficiency is increasing, we check the sign of E(t)E'(t) in the intervals determined by these critical points:

  1. For t<1.61t < 1.61
  2. For 1.61<t<24.861.61 < t < 24.86
  3. For t>24.86t > 24.86

We'll test the sign of E(t)E'(t) in these intervals.The results for E(t)E'(t) in the test intervals are:

  • E(t)<0E'(t) < 0 for t<1.61t < 1.61, indicating fuel efficiency is decreasing before t=1.61t = 1.61.
  • E(t)>0E'(t) > 0 for 1.61<t<24.861.61 < t < 24.86, indicating fuel efficiency is increasing between these values of tt.
  • E(t)<0E'(t) < 0 for t>24.86t > 24.86, indicating fuel efficiency is decreasing again after t=24.86t = 24.86.

Thus, fuel efficiency was improving between the years:

  • a=1.61a = 1.61 (which corresponds to 1971.61, or approximately 1972)
  • b=24.86b = 24.86 (which corresponds to 1994.86, or approximately 1995).

Final Answer:

  • Fuel efficiency was improving between 1972 and 1995.

Would you like further details or explanations on any part of this? Here are 5 related questions you might find interesting:

  1. How can we determine the maximum fuel efficiency within the interval?
  2. What is the significance of E(t)=0E(t) = 0 in this context?
  3. How can you solve more complex inequalities involving cubic or higher-degree functions?
  4. What other real-world applications might use a similar mathematical model?
  5. How does calculus help in optimizing real-life systems?

Tip: Quadratic inequalities can often be tested by checking the sign of the expression at various points in its intervals!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quadratic Inequalities
Function Analysis

Formulas

Derivative of a polynomial: d/dt [at^n] = n*at^(n-1)
Quadratic equation solution: ax^2 + bx + c = 0

Theorems

Quadratic Formula
Sign Test for Inequalities

Suitable Grade Level

Grades 11-12 or Early College