When a transistor radio is switched off, the current falls away according to the differential equation #(dI)/dt=-kI# where #k# Is a constant . If the current drops to 10% in the first second ,how long will it take to drop to 0.1% of its original value?
# 3 \ s#
It is a First Order Separable Ordinary Differential Equation, allowing us to "separate the variables" and gather terms to obtain:
which are composed of standard integrals, allowing us to integrate directly to obtain:
Furthermore, since the exponential is positive across the board, we can write:
Since we know that in the first second the current decreases to 10% of its starting value, we can calculate:
And:
As a result, we can write the answer as:
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The given differential equation is ( \frac{dI}{dt} = -kI ).
We are told that when ( t = 1 ), ( I = 0.1I_0 ), where ( I_0 ) is the original value of the current.
Using separation of variables and integrating both sides, we get:
[ \int \frac{1}{I} , dI = \int -k , dt ]
This simplifies to:
[ \ln|I| = -kt + C ]
Using the initial condition when ( t = 1 ) and ( I = 0.1I_0 ), we find the constant ( C ):
[ \ln|0.1I_0| = -k + C ]
[ C = \ln|0.1I_0| + k ]
Now, substitute ( C ) back into the equation:
[ \ln|I| = -kt + \ln|0.1I_0| + k ]
[ \ln\left|\frac{I}{0.1I_0}\right| = -kt + k ]
[ \ln\left|\frac{I}{0.1I_0}\right| = k(-t + 1) ]
[ \frac{I}{0.1I_0} = e^{k(-t + 1)} ]
[ I = 0.1I_0 e^{k(1 - t)} ]
To find when the current drops to ( 0.001I_0 ), set ( I = 0.001I_0 ):
[ 0.001I_0 = 0.1I_0 e^{k(1 - t)} ]
[ 0.001 = 0.1e^{k(1 - t)} ]
[ 0.01 = e^{k(1 - t)} ]
Taking the natural logarithm of both sides:
[ \ln(0.01) = k(1 - t) ]
[ -4.605 = k(1 - t) ]
[ -4.605/k = 1 - t ]
[ t = 1 - \frac{4.605}{k} ]
This gives the time it takes for the current to drop to ( 0.1% ) of its original value.
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To find the time it takes for the current to drop to 0.1% of its original value, we need to solve the given differential equation (\frac{dI}{dt} = -kI).
Given that the current drops to 10% of its original value in the first second, we can use this information to find the value of (k).
Let's denote the original current as (I_0). After one second, the current is (0.1I_0). Using the given differential equation, we can set up the following equation:
[\frac{dI}{dt} = -kI] [\int_{I_0}^{0.1I_0} \frac{1}{I} dI = \int_{0}^{1} -k dt] [\ln\left(\frac{0.1I_0}{I_0}\right) = -kt \Bigg|_0^1] [\ln(0.1) = -k(1) - (-k(0))] [\ln(0.1) = -k] [k = -\ln(0.1)]
Now, we can use the value of (k) to find the time it takes for the current to drop to 0.1% of its original value.
Let (t) represent this time. We want to find (t) when (I = 0.001I_0). Substituting into the differential equation:
[\frac{dI}{dt} = -kI] [\int_{I_0}^{0.001I_0} \frac{1}{I} dI = \int_{0}^{t} -k dt] [\ln\left(\frac{0.001I_0}{I_0}\right) = -k t] [\ln(0.001) = -k t] [t = \frac{\ln(0.001)}{-k}]
Substitute the value of (k) we found earlier:
[t = \frac{\ln(0.001)}{-(-\ln(0.1))}]
[t = \frac{\ln(0.001)}{\ln(0.1)}]
[t \approx \frac{-6.9078}{-2.3026}]
[t \approx 3]
So, it will take approximately 3 seconds for the current to drop to 0.1% of its original value.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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