A plane is flying horizontally at P at a height of 150 m above level ground when it begins its descent. P is 5 km, horizontally, from the point of touchdown O. The plane lands horizontally at O. Taking O as the origin, $(x, f(x))$ approximately describes the path of the plane's descent where $$f(x) = -0.0024x^3 + 0.018x^2 + cx + d,$$ and $-5 ext{ km} ext{ } < ext{ } x ext{ } < ext{ } 0$, and both $f(x)$ are measured in km. (a) Show that $d = 0$. (b) (i) Find the value of $f'(x)$, the derivative of $f(x)$, when $x = -4$. (ii) Use your answer to part (b) above to find the angle at which the plane is descending when it is 4 km from touchdown. Give your answer correct to the nearest degree. (c) Show that $(-2.5, 0.075)$ is the point of inflection of the curve $y = f(x)$. (d) (i) If $(x, y)$ is a point on the curve $y = f(x)$, verify that $(-x - 5, -y + 0.15)$ is also a point on $y = f(x)$. (ii) Find the image of $(-x - 5, -y + 0.15)$ under symmetry in the point of inflection.

Leaving Cert Mathematics Differentiation: A plane is flying horizontally a

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

Question 7

A plane is flying horizontally at P at a height of 150 m above level ground when it begins its descent. P is 5 km, horizontally, from the point of touchdown O. The p... show full transcript

Worked Solution & Example Answer: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

Step 1

Show that $d = 0$.

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Answer

Given the function $f(x) = -0.0024x^3 + 0.018x^2 + cx + d$ At point O, which corresponds to $x = 0$ , we have: $f(0) = d.$ Since the plane lands at point O horizontally, we set $f(0) = 0.$ Thus, substituting, we get:

\implies d = 0.$$

Step 2

Using the fact that P is the point $(-5, 0.15)$, or otherwise, show that $c = 0$.

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Substituting the coordinates of point P into the equation: $f(-5) = -0.0024(-5)^3 + 0.018(-5)^2 + c(-5) + d.$ Plugging in the known values: $0.15 = -0.0024(-125) + 0.018(25) + (-5c) + 0$ Calculating:

0.15 = 0.75 - 5c \ 5c = 0.75 - 0.15 \ 5c = 0.6 \ c = 0.12. $$

Step 3

Find the value of $f'(x)$, the derivative of $f(x)$, when $x = -4$.

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To find the derivative: $f'(x) = -0.0072x^2 + 0.036x$ Now substituting $x = -4$ :

= -0.0072(16) - 0.144\ = -0.1152 - 0.144\ = -0.2592.$$ Thus, $f'(-4) = -0.288$.

Step 4

Use your answer to part (b) above to find the angle at which the plane is descending when it is 4 km from touchdown.

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To find the angle of descent, we use: $tan(\theta) = f'(-4).$ Where: $\theta = tan^{-1}(-0.288)$ Calculating: $\theta = 178.3503^ ext{o}.$ Thus, the angle is approximately $\theta = 179^ ext{o}$ .

Step 5

Show that $(-2.5, 0.075)$ is the point of inflection of the curve $y = f(x)$.

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To find the point of inflection: We need to find $f''(x)$ : $f''(x) = -0.0144x + 0.036.$ Setting $f''(x) = 0$ :

\implies x = -2.5.$$ Now verifying the function value: $$f(-2.5) = -0.0024(-2.5)^3 + 0.018(-2.5)^2 + c(-2.5) + 0$$ Calculating for $(-2.5, 0.075)$, we have: $$f(-2.5) = -0.075 + 0.1125 = 0.075.$$

Step 6

If $(x, y)$ is a point on the curve $y = f(x)$, verify that $(-x - 5, -y + 0.15)$ is also a point on $y = f(x)$.

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Answer

Given the points relationship: $f(-x - 5) = -f(x) + 0.15.$ Substituting: $(-x - 5, -y + 0.15) \Rightarrow f(-x - 5) = f(x) + 0.15.$ Thus proving that $(-x-5, -y + 0.15)$ is indeed also a point on the curve.

Step 7

Find the image of $(-x - 5, -y + 0.15)$ under symmetry in the point of inflection.

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To find the image point $(x', y')$ , under symmetry: Using $(x', y') = \left( \frac{-x - 5 + x_{inflection}}{2}, \frac{-y + 0.15 + y_{inflection}}{2} \right)$ Given the point of inflection $(-2.5, 0.075)$ : $x' = \frac{-(-5) - 5 + (-2.5)}{2} = -2.5$ $y' = \frac{-(-y + 0.15) + 0.075}{2} = -0.075.$ Thus the final image point is $(-2.5, -0.075)$ .

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

Worked Solution & Example Answer: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

Show that $d = 0$.

Using the fact that P is the point $(-5, 0.15)$, or otherwise, show that $c = 0$.

Find the value of $f'(x)$, the derivative of $f(x)$, when $x = -4$.

Use your answer to part (b) above to find the angle at which the plane is descending when it is 4 km from touchdown.

Show that $(-2.5, 0.075)$ is the point of inflection of the curve $y = f(x)$.

If $(x, y)$ is a point on the curve $y = f(x)$, verify that $(-x - 5, -y + 0.15)$ is also a point on $y = f(x)$.

Find the image of $(-x - 5, -y + 0.15)$ under symmetry in the point of inflection.

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