Determining the number of gallons of fuel that the car used on the trip.

Question

My answer was (C), where I was thinking while I was solving how the unit could be Gallons only, but the correct answer is (A), I do not know why my answer is wrong and why (A) is the correct answer, could anyone help me please?

Answer 1

So $E(v(t))$ is $\dfrac{\text{miles}}{\text{gallon}}$, or in calculus terms, $\frac{dm}{dg}$. Your speed is $\frac{dm}{dt}$ (miles per hour). So to get $\frac{dg}{dt}$, the derivative you need, it is $\dfrac{\frac{dm}{dt}}{\frac{dg}{dt}}$, which is $\dfrac{v(t)}{E(v(t))}$.

Answer 2

So, think about it like this - you want the integral to give you the amount of gallons you use. That means the area under your curve is going to be the gallons you used in relation to the time you spent driving.

Just look at what each one is telling you in the equation itself:

A) $\frac{v(t)}{E(v(t))}$

Here right away we see that we are saying $\frac{\frac{miles}{hour}}{\frac{miles}{gallon}} = \frac{gallon*miles}{miles*hour} = \frac{gallon}{hour} $ which means we're looking at gallons per time unit (hours) which means when we integrate this function we will be able to say we used exactly this many gallons in this many hours. Of course, this is somewhat trivial if it's a linear function, but let's say you're constantly changing your speed - well that effects your mpg and so here we've taken your speed completely out of it because, honestly, we don't care at the end of the day about your speed: just how hungry your car was for that sweet, sweet Dino juice.

anyways, take a look at the other functions and see what you're really getting when you integrate them. ;)