Question

the value of one share of stock, in dollars, t hours after 12 p.m. is modeled by ( v(t) = 14 - 0.1t + 0.8t^2 ), while the total number of shares traded t hours after 12 p.m. is modeled by ( s(t) = 1235e^{0.04t} ). the total value of all traded shares t hours after 12 p.m. is given by ( f(t) = v(t) cdot s(t) ).
a. find ( f(t) = )

b. how fast is the total value of the traded shares changing at 5 p.m.? (round to the nearest cent.)

Explanation:

Part A

Step1: Identify the product rule

We have \( f(t)=v(t)\cdot s(t) \), so we use the product rule: \( (uv)' = u'v + uv' \), where \( u = v(t) \) and \( v = s(t) \).

Step2: Find derivatives of \( v(t) \) and \( s(t) \)

• For \( v(t)=14 - 0.1t + 0.8t^{2} \), the derivative \( v'(t)=\frac{d}{dt}(14 - 0.1t + 0.8t^{2})=- 0.1 + 1.6t \)
• For \( s(t)=1235e^{0.04t} \), the derivative \( s'(t)=\frac{d}{dt}(1235e^{0.04t})=1235\times0.04e^{0.04t}=49.4e^{0.04t} \)

Step3: Apply product rule

Using the product rule \( f'(t)=v'(t)s(t)+v(t)s'(t) \), substitute the functions:
\( f'(t)=(-0.1 + 1.6t)\times1235e^{0.04t}+(14 - 0.1t + 0.8t^{2})\times49.4e^{0.04t} \)

Step1: Determine the time \( t \)

5 p.m. is 5 hours after 12 p.m., so \( t = 5 \).

Step2: Substitute \( t = 5 \) into \( f'(t) \)

First, calculate each part:

• For \( (-0.1 + 1.6t)\cdot1235e^{0.04t} \) when \( t = 5 \):

\( (-0.1+1.6\times5)\times1235e^{0.04\times5}=(-0.1 + 8)\times1235e^{0.2}=7.9\times1235e^{0.2} \)
\( 7.9\times1235 = 9756.5 \), \( e^{0.2}\approx1.2214 \), so this part is \( 9756.5\times1.2214\approx11917.5 \)

• For \( (14 - 0.1t + 0.8t^{2})\cdot49.4e^{0.04t} \) when \( t = 5 \):

\( 14-0.1\times5 + 0.8\times5^{2}=14 - 0.5+20 = 33.5 \)
\( 33.5\times49.4e^{0.2}=33.5\times49.4\times1.2214 \)
\( 33.5\times49.4 = 1654.9 \), \( 1654.9\times1.2214\approx2021.3 \)

Step3: Sum the two parts

Add the two results: \( 11917.5+2021.3 = 13938.8 \) (rounded to the nearest cent)

Answer:

\( f'(t)=(-0.1 + 1.6t)\cdot1235e^{0.04t}+(14 - 0.1t + 0.8t^{2})\cdot49.4e^{0.04t} \)

Part B