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

Bytelearn · Accepted Answer

Differentiate function W(t): To find the instantaneous rate of change, we need to differentiate the function W(t) with respect to t. W(t) = 4sqrt(t) + ln(t+1) Let's differentiate it. dW/dt = d/dt [4sqrt(t)] + d/dt [ln(t+1)] dW/dt = 4 * (1/2) * t^(-1/2) + 1/(t+1) dW/dt = 2/t^(1/2) + 1/(t+1)Substitute t = 100: Now we substitute t = 100 into the derivative to find the rate of change at t = 100 seconds. dW/dt = 2/100^(1/2) + 1/(100+1) dW/dt = 2/10 + 1/101 dW/dt = 0.2 + 0.00990099Calculate rate of change: Let's add these up to get the instantaneous rate of change. dW/dt = 0.2 + 0.00990099 dW/dt ≈ 0.209901Round to two decimal places: We round this to two decimal places as per the options given. dW/dt ≈ 0.21

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