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: Find \( f'(t) \)

Step 1: Identify the product rule

Since \( f(t) = v(t) \cdot s(t) \), we use the product rule for differentiation: \( (uv)' = u'v + uv' \), where \( u = v(t) \) and \( v = s(t) \).

Step 2: Differentiate \( v(t) \)

Given \( v(t) = 14 - 0.1t + 0.8t^2 \), its derivative \( v'(t) = -0.1 + 1.6t \).

Step 3: Differentiate \( s(t) \)

Given \( s(t) = 1235e^{0.04t} \), its derivative \( s'(t) = 1235 \cdot 0.04e^{0.04t} = 49.4e^{0.04t} \).

Step 4: Apply the product rule

\( f'(t) = v'(t) \cdot s(t) + v(t) \cdot s'(t) \)
Substitute \( v'(t) \), \( s(t) \), \( v(t) \), and \( s'(t) \):
\( f'(t) = (-0.1 + 1.6t) \cdot 1235e^{0.04t} + (14 - 0.1t + 0.8t^2) \cdot 49.4e^{0.04t} \)

Step 1: Determine the value of \( t \) at 5 p.m.

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

Step 2: Substitute \( t = 5 \) into \( v(t) \), \( v'(t) \), \( s(t) \), \( s'(t) \)

• \( v(5) = 14 - 0.1(5) + 0.8(5)^2 = 14 - 0.5 + 20 = 33.5 \)
• \( v'(5) = -0.1 + 1.6(5) = -0.1 + 8 = 7.9 \)
• \( s(5) = 1235e^{0.04(5)} = 1235e^{0.2} \approx 1235 \cdot 1.221402758 = 1508.4323 \)
• \( s'(5) = 49.4e^{0.2} \approx 49.4 \cdot 1.221402758 \approx 60.3473 \)

Step 3: Substitute into \( f'(t) \)

Using \( f'(t) = v'(t)s(t) + v(t)s'(t) \):
\( f'(5) = 7.9 \cdot 1508.4323 + 33.5 \cdot 60.3473 \)
First term: \( 7.9 \cdot 1508.4323 \approx 11916.6152 \)
Second term: \( 33.5 \cdot 60.3473 \approx 2021.6346 \)
Sum: \( 11916.6152 + 2021.6346 \approx 13938.25 \)

Answer:

\( f'(t) = (-0.1 + 1.6t) \cdot 1235e^{0.04t} + (14 - 0.1t + 0.8t^2) \cdot 49.4e^{0.04t} \)

Part B: How fast is the total value changing at 5 p.m.?