Day 9: Christmas Ornament-Making

Starting off with the fact that $b \ = \ -\text{ln}(\frac{4}{A})$ , we can rearrange this to get $Ae^{-b} \ = \ 4$ . Let's look at one of the graphs, $f(x) \ = \ Ae^{(x \ - \ b)} \ + \ c$ . This becomes $f(x) \ = \ Ae^{-b}e^{x} \ + \ c \ \Rightarrow \ 4e^{x} \ + \ c$ . To find the mass, we simply need to find the surface area integral over the surface of the area under the curve, by integrating across the density function (to get the mass). We must calculate the value of:

$\displaystyle\int_{a_1}^{a_2} \ \displaystyle\int_{b_1}^{b_2} \ D(x, \ y) \ \text{dy} \ \text{dx}$

Since Doug wants his ornament to have a maximum height of $2$ units, and we know that the two graphs intersect at the x-axis, therefore that is where the maximum height will be. We then get $4e^0 \ + \ c \ = \ 2 \ \Rightarrow \ c \ = \ -2$ .

If we allow $x$ to vary from its maximum to minimum, we get $\text{ln}(\frac{1}{2}) \ \leq \ x \ \leq \ 0$ , and setting the $y$ bounds accordingly, we get: $0 \ \leq \ 4e^x \ - \ 2$ . The only step that remains is solving the integral:

$\displaystyle\int_{-0.69}^{0} \ \displaystyle\int_{0}^{4e^x \ - \ 2} \ x^2 \ \text{dy} \ \text{dx} \ \Rightarrow \ \displaystyle\int_{-0.69}^{0} \ (4e^x \ - \ 2) \ x^2 \ \text{dx} \ \Rightarrow \ 4\displaystyle\int_{-0.69}^{0} \ e^x \ x^2 \ \text{dx} \ - \ 2 \displaystyle\int_{-0.69}^{0} \ x^2 \ \text{dx}$

Solving this integral, we get the value $0.445$ . Since this shape is symmetric, and the density function is always positive along with the Christmas tree shape being above the x-axis, we can multiply the mass by two to get a final answer of $0.890$ .