Question

the ozone level (in parts per billion) on a summer day in a certain metropolitan area is given by $p(t)=100 + 17t - t^{2}$, where $t$ is time in hours and $t = 0$ corresponds to 9 am.
(a) use the four - step process to find $p^{prime}(t)$.
(b) find $p(4)$ and $p^{prime}(4)$.
(a) use the four - step process to find $p^{prime}(t)$.
$p^{prime}(t)=square$

Explanation:

Step1: Recall the four - step process for the derivative

The four - step process for finding the derivative of a function $y = f(x)$ is based on the limit definition $f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Here $P(t)=100 + 17t-t^{2}$. First, find $P(t + h)$:
$P(t + h)=100+17(t + h)-(t + h)^{2}=100+17t+17h-(t^{2}+2th + h^{2})=100+17t+17h - t^{2}-2th - h^{2}$

Step2: Calculate $P(t + h)-P(t)$

$P(t + h)-P(t)=(100+17t+17h - t^{2}-2th - h^{2})-(100 + 17t-t^{2})$
$=100+17t+17h - t^{2}-2th - h^{2}-100 - 17t + t^{2}=17h-2th - h^{2}$

Step3: Calculate $\frac{P(t + h)-P(t)}{h}$

$\frac{P(t + h)-P(t)}{h}=\frac{17h-2th - h^{2}}{h}=17 - 2t - h$

Step4: Find the limit as $h

ightarrow0$
$P^\prime(t)=\lim_{h
ightarrow0}\frac{P(t + h)-P(t)}{h}=\lim_{h
ightarrow0}(17 - 2t - h)=17-2t$

Step5: Find $P(4)$

Substitute $t = 4$ into $P(t)$:
$P(4)=100+17\times4-4^{2}=100 + 68-16=152$

Step6: Find $P^\prime(4)$

Substitute $t = 4$ into $P^\prime(t)$:
$P^\prime(4)=17-2\times4=17 - 8 = 9$

Answer:

(A) $P^\prime(t)=17 - 2t$
(B) $P(4)=152$, $P^\prime(4)=9$