Question

A plane takes off from an airport which is at sea level. The aircraft’s altitude in metres at a time of t minutes is modelled by the equation

h = 5400 ln(t + 1)

a) What is the height of the plane after five minutes?

b) What is the rate of climb of the plane after five minutes?

c) Using differentiation, show the curve y = h(t) has no stationary points, and hence comment on how realistic the model would be for large t.

Answer 1

a)

$$h(5)=5400\ln(5+1)=5400\ln(6)\approx9675.5 (m)$$

b)

$$h'(t)=\dfrac{5400}{t+1}$$

$$h'(5)=\dfrac{5400}{5+1}=900 (m/min)$$

c)

$$h(t)=5400\ln(t+1),t\geq0$$

$$h'(t)=\dfrac{5400}{t+1}$$

Since $h'(t)>0$ for $t\geq0,$ the curve y = h(t) has no stationary points.

The height of the plane increases to infinity. as $t\to \infin.$

How does the plane land?