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

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

1 Answer
Oct 2, 2017

Delta% = -23.66%

Explanation:

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%