Solving a Problem: Have I Made a Mistake or Are the Solutions Wrong?

Darkmisc · Jul 28, 2022

Hi everyone

To solve the below problem, I assumed the affected area was 2x2 minus the definite integral of the given function between 2 and 4.

I then equated the answer for that with the given function to solve for a, b and c.

I don't know why the solutions give b as 2ln5.

Have I made a mistake, or are the solutions wrong?

Thanks

Handwritten calculus working on lined paper: A = 4 - integral from 2 to 4 of (1/2)e^(x/2) - 1/2 dx, then antiderivative e^(x/2) - x/2 evaluated from 2 to 4, giving 4 - [(e^2 - 2) - (e - 1)] = 5 - e^2 + e.

Handwritten derivation of a definite integral from a to b of c minus one half e to the power x over 2, giving [cx - e^(x/2)] from a to b, then (bc - e^(b/2)) - (ac - e^(a/2)), leading to values b = 4, a = 2, c = 5/2.

Diagram of a fire front curve rising from the town at (0, 0) across a grid of x and g(x) axes, with a farm photo at x from 2 to 4 and a compass rose showing north, south, east and west.

Typed text stating that from first sighting until a fire is fully extinguished, the fire front position is given by g(x) = (1/(2k)) e^(x/(2k)) - 1/2, where k is a parameter with k ≥ 1/2.

Handwritten working on a printed exam question: a = 2, b = 2 log_e(5), c = 5/2 or 2.5, beside the printed integral from a to b of (c - (1/2)e^(x/2)) dx.

andrewkirk · Jul 29, 2022

You are trying to work out a, b and c by solving equations. That cannot work, as you have three unknowns and only one equation. Instead you set the values a and b from first principles, as the lower and upper bounds of x at which the fire front intersects the farm. Given the fire front equation is ##f(x) = \frac12 e^{\frac x2}-\frac12## and the farm is ##[2,4]\times [0,2]## we see that the intersection points are ##(2,e^\frac12)## and ##(b,2)##, the second point being where the fire front intersects the line ##y=2##. That second point gives us the equation
$$2 = f(b)=\frac12 e^\frac b2-\frac12$$
which we solve to get
$$b=2\log 5$$
Now that you know ##a## and ##b## you can solve the equation to find ##c##.

Delta2 · Jul 30, 2022

A small correction to @andrewkirk post ,the first point of intersection is ##(2,f(2))=(2,\frac{e-1}{2})##.

Delta2 · Jul 30, 2022

I also think that the "4=(4-2)x(2-0)" in your equation for A shouldn't be 4 but instead ##(b-2)\times(2-0)=2(b-2)##, hard to explain with words without making a scheme (I am really bad in making schemes).

Darkmisc · Jul 30, 2022

Yeah, I drew the diagram for myself wrong. I assumed the fire front would touch the right edge of the property (which it didn't).

Solving a Problem: Have I Made a Mistake or Are the Solutions Wrong?

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