A pool is being filled. The following function gives the pool's water level (in centimeters) after t seconds: W(t)=4sqrtt+ln(t+1) What is the instantaneous rate of change of the pool's water level after 100 seconds? Choose 1 answer: (A) 0.21 centimeters (B) 0.21 centimeters per second (C) 44.6 centimeters (D) 44.6 centimeters per second

Question

A pool is being filled. The following function gives the pool's water level (in centimeters) after  t seconds:  W(t)=4sqrtt+ln(t+1) What is the instantaneous rate of change of the pool's water level after 100 seconds? Choose 1 answer: (A) 0.21 centimeters (B) 0.21 centimeters per second (C) 44.6 centimeters (D) 44.6 centimeters per second

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Differentiate Function: To find the instantaneous rate of change, we need to differentiate the function W(t) with respect to t.Evaluate Derivative at t=100: Differentiate W(t) = 4sqrt(t) + ln(t+1). The derivative of 4sqrt(t) is 2/sqrt(t) and the derivative of ln(t+1) is 1/(t+1). So, W'(t) = 2/sqrt(t) + 1/(t+1).Calculate Instantaneous Rate: Now we need to evaluate W'(t) at t = 100 seconds to find the instantaneous rate of change at that moment.Plug t=100 into W'(t): Plug t = 100 into W'(t) to get W'(100) = 2/sqrt(100) + 1/(100+1).Add Results: Calculate W'(100) = 2/10 + 1/101. W'(100) = 0.2 + 0.00990099.Round to Two Decimal Places: Add 0.2 and 0.00990099 to get the instantaneous rate of change. W'(100) = 0.20990099.Determine Units: Round the answer to two decimal places since the choices are given in two decimal places. W'(100) ≈ 0.21.Correct Answer: The units for the rate of change should be in centimeters per second because it's the change in water level over time.The correct answer is 0.21 centimeters per second, which corresponds to choice (B).

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