A college student develops an engine that is believed to meet all state standards for emission control. The new engine’s emission rate is given by E(t) = 2t^2, where E(t) is the em... A college student develops an engine that is believed to meet all state standards for emission control. The new engine’s emission rate is given by E(t) = 2t^2, where E(t) is the emissions in billions of population particulates per year, at time t, in years. The emission rate of a conventional engine is given by C(t) = 9 + t^2. (a) At what point in time will the emission rates be the same? (b) What reduction in emissions results from using the student’s engine? (Use integral) Read more

Steps to Solve

1. Set the emission rates equal To find the point in time when emissions from both engines are the same, we set $E(t)$ equal to $C(t)$: $$ 2t^2 = 9 + t^2 $$

2. Rearrange the equation Subtract $t^2$ from both sides: $$ 2t^2 - t^2 = 9 $$ This simplifies to: $$ t^2 = 9 $$

3. Solve for (t) Taking the square root of both sides: $$ t = \sqrt{9} \implies t = 3 \text{ (considering only the positive root since time cannot be negative)} $$

4. Calculate emissions using integration To find the total emissions from both engines until (t = 3), we integrate the emission functions from (t = 0) to (t = 3).

For the student's engine: $$ \int_0^3 E(t) , dt = \int_0^3 2t^2 , dt $$ Calculating gives: $$ = 2 \cdot \left[\frac{t^3}{3}\right]_0^3 = 2 \cdot \frac{27}{3} = 18 $$

For the conventional engine: $$ \int_0^3 C(t) , dt = \int_0^3 (9 + t^2) , dt $$ Calculating gives: $$ = \left[9t + \frac{t^3}{3}\right]_0^3 = 9(3) + \frac{27}{3} = 27 + 9 = 36 $$

5. Calculate the reduction in emissions The reduction in emissions when using the student’s engine compared to the conventional engine is: $$ \text{Reduction} = 36 - 18 = 18 $$