Math  /  Calculus

QuestionA model helicopter takes off from a point OO at time t=0 st=0 \mathrm{~s} and moves vertically so that its height y cmy \mathrm{~cm}, above OO after time tt seconds is given by y=14t426t2+96t,0t4y=\frac{1}{4} t^{4}-26 t^{2}+96 t, \quad 0 \leq t \leq 4 a) Find (i) dydt\frac{d y}{d t} (ii) d2ydt2\frac{d^{2} y}{d t^{2}} b) Verify that yy has a stationary value when t=2t=2 and determine whether this stationary value is a maximum or a minimum value. c) Find the rate of change of yy with respect to tt when t=1t=1. d) Determine whether the height of the helicopter is increasing or decreasing at the instant when t=3t=3.

Studdy Solution
(a) (i) dydt=t352t+96\frac{dy}{dt} = t^3 - 52t + 96 (ii) d2ydt2=3t252\frac{d^2y}{dt^2} = 3t^2 - 52 (b) The stationary value at t=2t=2 is a **maximum**. (c) The rate of change of yy at t=1t=1 is 45 cm/s\textbf{45 cm/s}. (d) The height is **decreasing** at t=3t=3.

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