V(t) = -5sin(3t) with V in volts and t in seconds. Compute the rate of change of V at the instant t = 1 second. Your answer must be accurate to one decimal place. NOTE: The best way to do this is to plug u = 1 into your secant slope formula. Too bad that's impossible. You will have to use your table of secant slopes to estimate the answer. V'(1) = .83 V/sec
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Step 1: The rate of change of V at the instant t=1 second can be found by taking the derivative of V(t) with respect to t. Show more…
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As in the previous problem, electric potential in a circuit is given by V(t) = − 5 sin(3t) with V in volts and t in seconds. Compute the rate of change of V at the instant t = 1 second. Your answer must be accurate to one decimal place. NOTE: The best way to do this is to plug u = 1 into your secant slope formula. Too bad that's impossible. You will have to use your table of secant slopes to estimate the answer. V '(1) =
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