A-Level Edexcel Maths Pure Vectors in 2 Dimensions: The current, I amps, in an elect

The current, I amps, in an electric circuit at time t seconds is given by I = 16 - 16(0.5)^t, t >= 0 Use differentiation to find the value of $ \frac{dI}{dt} $ when t = 3 - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 6

Question 4

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The current, I amps, in an electric circuit at time t seconds is given by I = 16 - 16(0.5)^t, t >= 0 Use differentiation to find the value of $ \frac{dI}{dt} $ ... show full transcript

Worked Solution & Example Answer:The current, I amps, in an electric circuit at time t seconds is given by I = 16 - 16(0.5)^t, t >= 0 Use differentiation to find the value of $ \frac{dI}{dt} $ when t = 3 - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 6

Step 1

Use differentiation to find $ \frac{dI}{dt} $

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Answer

To find ( \frac{dI}{dt} ), we start by differentiating the given equation:

[ I = 16 - 16(0.5)^t ]

Differentiating with respect to t:

[ \frac{dI}{dt} = -16 \cdot \ln(0.5) \cdot (0.5)^t ]

Step 2

Evaluate $ \frac{dI}{dt} $ at t = 3

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Answer

Now, substituting t = 3 into our derivative:

[ \frac{dI}{dt} = -16 \cdot \ln(0.5) \cdot (0.5)^3 ]

Calculating ( (0.5)^3 ):

[ (0.5)^3 = \frac{1}{8} ]

So,

[ \frac{dI}{dt} = -16 \cdot \ln(0.5) \cdot \frac{1}{8} = -2 \ln(0.5) ]

Step 3

Express the answer in the form $ \ln a $

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Answer

Since ( \ln(0.5) ) can be rewritten, we have:

[ \frac{dI}{dt} = -2 \ln(0.5) = \ln((0.5)^{-2}) = \ln(\frac{1}{(0.5)^2}) = \ln(4) ]

Thus, the answer is in the form ( \ln a ), where a = 4.

The current, I amps, in an electric circuit at time t seconds is given by I = 16 - 16(0.5)^t, t >= 0 Use differentiation to find the value of \( \frac{dI}{dt} \) when t = 3 - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 6

Worked Solution & Example Answer:The current, I amps, in an electric circuit at time t seconds is given by I = 16 - 16(0.5)^t, t >= 0 Use differentiation to find the value of \( \frac{dI}{dt} \) when t = 3 - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 6

Use differentiation to find \( \frac{dI}{dt} \)

Evaluate \( \frac{dI}{dt} \) at t = 3

Express the answer in the form \( \ln a \)

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