11. -/1 points details my notes tanapcalc10 4.2.090. flight of a rocket the altitude (in feet) of a rocket t…

11. -/1 points details my notes tanapcalc10 4.2.090. flight of a rocket the altitude (in feet) of a rocket t s into flight is given by s = f(t)= -t³ + 48t² + 570t + 6 (t≥0). find the point of inflection of the function f. (t, s) = () what is the maximum velocity (in ft/s) attained by the rocket? ft/s

11. -/1 points details my notes tanapcalc10 4.2.090. flight of a rocket the altitude (in feet) of a rocket t s into flight is given by s = f(t)= -t³ + 48t² + 570t + 6 (t≥0). find the point of inflection of the function f. (t, s) = () what is the maximum velocity (in ft/s) attained by the rocket? ft/s

Answer

Explanation:

Step1: Find the first - derivative

The position function is $s = f(t)=-t^{3}+48t^{2}+570t + 6$. Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, the velocity function $v(t)=f^\prime(t)=-3t^{2}+96t + 570$.

Step2: Find the second - derivative

Differentiate $v(t)$ with respect to $t$. Using the power rule again, $a(t)=f^{\prime\prime}(t)=-6t + 96$.

Step3: Find the inflection point

Set $f^{\prime\prime}(t) = 0$. So, $-6t+96 = 0$. Solving for $t$: [ \begin{align*} -6t&=-96\ t& = 16 \end{align*} ] Substitute $t = 16$ into the position function $s=f(16)=-(16)^{3}+48\times(16)^{2}+570\times16 + 6=-4096+12288+9120 + 6=17318$. So the inflection point $(t,s)=(16,17318)$.

Step4: Find the maximum of the velocity function

Since $v(t)=-3t^{2}+96t + 570$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-3$, $b = 96$, $c = 570$. The vertex of a quadratic function $y = ax^{2}+bx + c$ occurs at $x=-\frac{b}{2a}$. For $v(t)$, the time at which the velocity is maximum is $t=-\frac{96}{2\times(-3)} = 16$. Substitute $t = 16$ into the velocity function $v(16)=-3\times(16)^{2}+96\times16+570=-768 + 1536+570=1338$.

Answer:

$(t,s)=(16,17318)$ $1338$