Question

attempt 1: 10 attempts remaining. the value, in dollars, of a stock t hours into a day of trading is modeled by v(t) = √(1400t + 5000). find the rate at which the stock’s value is increasing or decreasing after 2 hours. (round your answer to the nearest cent.) the stock’s value is ? by ? after 2 hours. submit answer next item

Explanation:

Step1: Rewrite the function

Rewrite $V(t)=\sqrt{1400t + 5000}=(1400t + 5000)^{\frac{1}{2}}$.

Step2: Differentiate using chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=1400t + 5000$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dt}=1400$. So $V^\prime(t)=\frac{1}{2}(1400t + 5000)^{-\frac{1}{2}}\times1400=\frac{1400}{2\sqrt{1400t + 5000}}=\frac{700}{\sqrt{1400t + 5000}}$.

Step3: Evaluate at $t = 2$

Substitute $t = 2$ into $V^\prime(t)$. We have $V^\prime(2)=\frac{700}{\sqrt{1400\times2+5000}}=\frac{700}{\sqrt{2800 + 5000}}=\frac{700}{\sqrt{7800}}\approx\frac{700}{88.32}\approx7.93$.

Answer:

The stock's value is increasing by $\$7.93$ after 2 hours.