By what percent does the voltage decrease each second? I'm so close to the answer help!?

Question

For A. I got #7e^(-0.27*4)#=2.4

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Answer 1 · 2017-10-02T17:59:00

#Delta% = -23.66%#

Given: #V(t) = 7e^(-0.27(t))#

Starting at a time any time #t#, you want the change, #(Delta)#, in percent one second later.

Let's take a moment to recall the definition of #Delta%#:

#Delta% = 100("NewValue" - "OldValue")/"OldValue"#

This can be written as:

#Delta% = 100("NewValue"/"OldValue" - 1)#

In our case, #"NewValue" = V(t+1)# and #"OldValue" = V(t)#

#Delta% = 100((V(t+1))/(V(t)) - 1)#

Substitute in the equivalents for the functions:

#Delta% = 100((7e^(-0.27(t+1)))/(7e^(-0.27(t))) - 1)#

Distribute the #-0.27#:

#Delta% = 100((7e^(-0.27(t)-0.27))/(7e^(-0.27(t))) - 1)#

We know that addition of a negative in the exponent is the same as multiplication by base raised to that negative exponent:

#Delta% = 100((7e^(-0.27(t))e^-0.27)/(7e^(-0.27(t))) - 1)#

Cancel the common factors:

#Delta% = 100((cancel(7e^(-0.27(t)))e^-0.27)/(cancel(7e^(-0.27(t)))) - 1)#

With the cancelled factors removed:

#Delta% = 100(e^-0.27 - 1)#

Rounded to two decimal places:

#Delta% = -23.66%#