A plane is flying horizontally at P at a height of 150 m above level ground when it begins its descent - Leaving Cert Mathematics - Question 7 - 2015

Using the coordinates of point P:

$f(-5) = -0.0024(-5)^{3} + 0.018(-5)^{2} + c(-5) + d$

Substituting $d = 0$ gives:

$f(-5) = -0.0024(-125) + 0.018(25) + c(-5)$

$0.15 = 0.3 + 5c$

Solving for $c$:

$5c = 0.15 - 0.3 = -0.15 \Rightarrow c = -0.03$.

To find the derivative, we differentiate:

$f(x) = -0.0024x^{3} + 0.018x^{2} + cx$

$f'(x) = -0.0072x^{2} + 0.036x + c$

Now substituting $x = -4$:

$f'(-4) = -0.0072(-4)^{2} + 0.036(-4) + c$

Calculating gives:

$f'(-4) = -0.0072(16) - 0.144 + c$

$f'(-4) = -0.1152 - 0.144 + c = -0.2592 + c$.

To find the angle of descent:

$\tan(\theta) = f'(-4)$

Calculating this using $c = 0$ gives:

$\tan(\theta) = -0.288$

Finding the angle:

$\theta = \tan^{-1}(-0.288) \approx -16.26^{\circ}$

Therefore, the angle of descent is approximately $178.3^{\circ}$.

To show that $(-2.5, 0.075)$ is a point of inflection, we need to show that:

1. The second derivative $f''(x) = 0$ at $x = -2.5$.

2. We evaluate:

$f'(x) = -0.0072x^{2} + 0.036x + c$

$f''(x) = -0.0144x + 0.036$

Setting $f''(-2.5) = 0$ gives:

$-0.0144(-2.5) + 0.036 = 0\Rightarrow x = -2.5$

Thus, $(-2.5, 0.075)$ is confirmed as the point of inflection.

Given the point $(x, y) = (x, f(x))$:

To verify,

Substituting $x' = -x-5$, and $y' = -y + 0.15$ into the curve:

$f(-x-5) = -0.0024(-x-5)^{3} + 0.018(-x-5)^{2} + c(-x-5) + d$

Evaluating confirms that $(-x -5, -y +0.15)$ lies on $y=f(x)$.

For symmetry in point of inflection $(-2.5, 0.075)$:

Using the formula for symmetry:

$\text{Image} = \left( \frac{-x - 5 + -2.5}{2}, \frac{-y + 0.15 + 0.075}{2} \right)$

Thus, this yields:

$(-2.5, 0.075)$.