Question

the function $f(t)=ln(t + 1)-\frac{t^{2}}{7}+40$ models the amount of water in a container, in gallons, between time $t = 1$ minutes and $t = 14$ minutes. what is the rate of change of the water at $t = 13$ minutes? show your work here hint: to add an exponent ($x^{y}$), type \exponent\ or press \^\

Explanation:

Step1: Find derivative of $f(t)$

The rate of change is given by the derivative $f'(t)$. Using differentiation rules:

• Derivative of $\ln(t+1)$ is $\frac{1}{t+1}$
• Derivative of $-\frac{t^2}{7}$ is $-\frac{2t}{7}$
• Derivative of constant 40 is 0.

$$f'(t) = \frac{1}{t+1} - \frac{2t}{7}$$

Step2: Substitute $t=13$ into $f'(t)$

Plug $t=13$ into the derivative:
$$f'(13) = \frac{1}{13+1} - \frac{2(13)}{7}$$

Step3: Simplify the expression

Calculate each term:
$\frac{1}{14} - \frac{26}{7} = \frac{1}{14} - \frac{52}{14}$
$$f'(13) = \frac{1-52}{14} = -\frac{51}{14}$$

Answer:

$-\frac{51}{14}$ gallons per minute (or approximately -3.64 gallons per minute)