Question

b. use the first principles $f(x)=lim_{x
ightarrow h}\frac{f(x + h)-f(x)}{h}$ to show that the derivative of $f(x)=x^{2}-2x$ is $f(x)=2x - 2$. sl = 3
c. sketch the derivative graph of $g(x)$.
d. find the coordinates of the stationary points on the curve $y=x^{3}-6x^{2}$. sl = 2
e. the first derivative of the parametric equations $x = 4t^{2}$ and $y = 8t$ is $\frac{dy}{dx}=\frac{1}{t}$. find the second derivative $\frac{d^{2}y}{dx^{2}}$. sl = 3
f. a huge sky - rocket is to be launched to mark the beginning of the new year. the height of the sky - rocket in meters is given by $h = 28t-2t^{2}$ where $t$ is the time in seconds after the sky - rocket is launched. use the velocity equation of the sky - rocket to find the maximum height to which it reached.
g. the height in meters of a bullet fired into the air after $t$ seconds is given by $h(t)=200t(10 - t)$ m. find the maximum height reached by the bullet. sl = 3

Explanation:

Step1: Find derivative using first - principles for part b

Given $f(x)=x^{2}-2x$, then $f(x + h)=(x + h)^{2}-2(x + h)=x^{2}+2xh+h^{2}-2x - 2h$.
\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}\frac{(x^{2}+2xh+h^{2}-2x - 2h)-(x^{2}-2x)}{h}\\ &=\lim_{h ightarrow0}\frac{x^{2}+2xh+h^{2}-2x - 2h - x^{2}+2x}{h}\\ &=\lim_{h ightarrow0}\frac{2xh+h^{2}-2h}{h}\\ &=\lim_{h ightarrow0}(2x + h-2)\\ &=2x-2 \end{align*}$$

\]

Step2: Find stationary points for part d

For the curve $y = x^{3}-6x^{2}$, we first find the derivative $y^{\prime}=3x^{2}-12x$.
Set $y^{\prime}=0$, so $3x^{2}-12x = 0$. Factor out $3x$: $3x(x - 4)=0$.
This gives $x = 0$ or $x = 4$.
When $x = 0$, $y=0^{3}-6\times0^{2}=0$. When $x = 4$, $y=4^{3}-6\times4^{2}=64 - 96=-32$.
The stationary points are $(0,0)$ and $(4,-32)$.

Step3: Find second - derivative for part e

We know $\frac{dy}{dx}=\frac{1}{t}$, and by the chain - rule for parametric equations $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$.
First, $\frac{d}{dt}(\frac{1}{t})=-\frac{1}{t^{2}}$, and $\frac{dx}{dt}=8t$.
So $\frac{d^{2}y}{dx^{2}}=\frac{-\frac{1}{t^{2}}}{8t}=-\frac{1}{8t^{3}}$.

Step4: Find maximum height for part f

The height of the sky - rocket is $h = 28t-2t^{2}$. The velocity $v=h^{\prime}=28 - 4t$.
At the maximum height, $v = 0$. So $28 - 4t=0$, which gives $t = 7$.
Substitute $t = 7$ into the height equation: $h(7)=28\times7-2\times7^{2}=196 - 98 = 98$ meters.

Step5: Find maximum height for part g

First expand $h(t)=200t(10 - t)=2000t-200t^{2}$.
The derivative $h^{\prime}(t)=2000 - 400t$.
Set $h^{\prime}(t)=0$, so $2000 - 400t=0$, which gives $t = 5$.
Substitute $t = 5$ into the height equation: $h(5)=200\times5\times(10 - 5)=5000$ meters.

Answer:

b. Shown above.
d. $(0,0)$ and $(4,-32)$
e. $-\frac{1}{8t^{3}}$
f. 98 meters
g. 5000 meters