Q.89 - Calculus

Question

For Exercises 89 and 90, suppose that Ian is a climber who is planning a path over a glacier in Canada’s Icefield Range. Glaciers are a little like gelatin: They tend to form cracks (crevasses) when their surfaces are concave down. Cracks close up and travel is easy when they are concave up, as shown in the figure 

Using a map, Ian approximates the elevation of the glacier on a line that runs up through its center as 

$\mathrm{h}\left(\mathrm{x}\right)=1.2+.0095\mathrm{x}+0.037{\mathrm{x}}^2-0.0072{\mathrm{x}}^3+0.00046{\mathrm{x}}^4,$ where both  $\mathrm{x}$ and $\mathrm{h}\left(\mathrm{x}\right)$ are measured in miles. Find the areas of the glacier where it is concave down and hence where Ian will need to move away from the center to avoid the crevasses.

Step-by-Step Solution

Answer

The elevation function does not approach a concave down area for any $\mathrm{x}>0.$

Step 1: Given Information

The elevation of a glacier with respect to a line running through its centre can be determined using the $\mathrm{h}\left(\mathrm{x}\right)$formula.

$\mathrm{h}\left(\mathrm{x}\right)=1.2+0.0095\mathrm{x}+0.37{\mathrm{x}}^2-0.0072{\mathrm{x}}^{33}+0.00046{\mathrm{x}}^4$

Step 2: Calculations

The second derivative should be determined to determine the function's concavity. The first derivative is calculated as the first step in the procedure.

The second derivative is obtained by differentiating the first derivative.

$\begin{array}{l}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{h}\text{'}\left(\mathrm{x}\right)\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(0.0095+0.074\mathrm{x}-0.0216{\mathrm{x}}^2+0.00184{\mathrm{x}}^3\right)\\ =0+0.074\times 1-0.0216\times 2\mathrm{x}+0.00184\times 3{\mathrm{x}}^2\\ \mathrm{h}"\left(\mathrm{x}\right)==0.074-0.0432\mathrm{x}+0.0562{\mathrm{x}}^2\end{array}$

Then you should find the x interval where the second derivative is negative. The function does not approach a concave down area since the quadratic polynomial derived as the second derivative has no root and does not accept any negative values for any $\mathrm{x}$.