the amount of water passing through a particular dam in a day is given by $f(t)=-3.5t^{2}+78t + 900$ million…

the amount of water passing through a particular dam in a day is given by $f(t)=-3.5t^{2}+78t + 900$ million liters, where $1leq tleq60$ and $t$ is measured in days.\na. find the derivative function $\frac{df}{dt}$.\n$\frac{df}{dt}=square$\nb. at what rate was the amount of water passing through the dam changing on day 30?\ndaily water flowing through the dam was doing so at a rate of $square$ million liters per day.

the amount of water passing through a particular dam in a day is given by $f(t)=-3.5t^{2}+78t + 900$ million liters, where $1leq tleq60$ and $t$ is measured in days.\na. find the derivative function $\frac{df}{dt}$.\n$\frac{df}{dt}=square$\nb. at what rate was the amount of water passing through the dam changing on day 30?\ndaily water flowing through the dam was doing so at a rate of $square$ million liters per day.

Answer

Explanation:

Step1: Apply power - rule for differentiation

The power - rule states that if $y = ax^n$, then $\frac{dy}{dx}=nax^{n - 1}$. For $f(t)=-3.5t^{2}+78t + 900$, the derivative of $-3.5t^{2}$ is $2\times(-3.5)t^{2 - 1}=-7t$, the derivative of $78t$ is $78\times1\times t^{1 - 1}=78$, and the derivative of the constant $900$ is $0$. So, $\frac{df}{dt}=-7t + 78$.

Step2: Evaluate the derivative at $t = 30$

Substitute $t = 30$ into $\frac{df}{dt}$. $\frac{df}{dt}\big|{t = 30}=-7\times30+78$. $\frac{df}{dt}\big|{t = 30}=-210 + 78=-132$.

Answer:

a. $\frac{df}{dt}=-7t + 78$ b. $-132$