the value of one share of stock, in dollars, $t$ hours after 12 p.m. is modeled by $v(t)=20 +…

the value of one share of stock, in dollars, $t$ hours after 12 p.m. is modeled by $v(t)=20 + 0.5t+0.7t^{2}$, while the total number of shares traded $t$ hours after 12 p.m. is modeled by $s(t)=1445e^{0.06t}$. the total value of all traded shares $t$ hours after 12 p.m. is given by $f(t)=v(t)cdot s(t)$. how fast is the total value of the traded shares changing at 4 p.m.? (round to the nearest cent.)

the value of one share of stock, in dollars, $t$ hours after 12 p.m. is modeled by $v(t)=20 + 0.5t+0.7t^{2}$, while the total number of shares traded $t$ hours after 12 p.m. is modeled by $s(t)=1445e^{0.06t}$. the total value of all traded shares $t$ hours after 12 p.m. is given by $f(t)=v(t)cdot s(t)$. how fast is the total value of the traded shares changing at 4 p.m.? (round to the nearest cent.)

Answer

Explanation:

Step1: Find the product - rule formula

The product - rule states that if $f(t)=v(t)\cdot s(t)$, then $f^\prime(t)=v^\prime(t)s(t)+v(t)s^\prime(t)$.

Step2: Differentiate $v(t)$

Given $v(t)=20 + 0.5t+0.7t^{2}$, then $v^\prime(t)=\frac{d}{dt}(20)+\frac{d}{dt}(0.5t)+\frac{d}{dt}(0.7t^{2})=0.5 + 1.4t$.

Step3: Differentiate $s(t)$

Given $s(t)=1445e^{0.06t}$, then $s^\prime(t)=1445\times0.06e^{0.06t}=86.7e^{0.06t}$.

Step4: Calculate $v(t)$, $v^\prime(t)$, $s(t)$ and $s^\prime(t)$ at $t = 4$

For $v(t)$: $v(4)=20+0.5\times4 + 0.7\times4^{2}=20 + 2+0.7\times16=20 + 2+11.2 = 33.2$. For $v^\prime(t)$: $v^\prime(4)=0.5+1.4\times4=0.5 + 5.6 = 6.1$. For $s(t)$: $s(4)=1445e^{0.06\times4}=1445e^{0.24}\approx1445\times1.271249\approx1836.955$. For $s^\prime(t)$: $s^\prime(4)=86.7e^{0.06\times4}=86.7e^{0.24}\approx86.7\times1.271249\approx109.117$.

Step5: Calculate $f^\prime(4)$

$f^\prime(4)=v^\prime(4)s(4)+v(4)s^\prime(4)$ $f^\prime(4)=6.1\times1836.955+33.2\times109.117$ $f^\prime(4)=11205.4255+3622.6844$ $f^\prime(4)=14828.11$

Answer:

$$14828.11$