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The quantity of charge of $ Q $ in coulombs (C) that has passed through a point in a wire up to time $ t $ (measured in seconds) is given by $ Q(t) = t^3 - 2t^2 + 6t + 2. $ Find the current when (a) $ t = 0.5 s $ and (b) $ t = 1 s. $ [See Example 3. The unit of current is an ampere ( 1 A = 1 C/s).] At what time is the current lowest?

(a) $Q^{\prime}(0.5)=3(0.5)^{2}-4(0.5)+6=4.75 \mathrm{A}$(b) $Q^{\prime}(1)=3(1)^{2}-4(1)+6=5 \mathrm{A}$

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Amrita B.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 7

Rates of Change in the Natural and Social Sciences

Derivatives

Differentiation

Campbell University

Oregon State University

Idaho State University

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here we have an equation that represents the charge at a point on a wire as a function of time, and we're interested in the current. And we learned from the lesson in the book that the current is the rate of change of the charge, the derivative. So if we take the derivative of this function, we get three t squared, minus 40 plus six, and we want to find the derivative at various points in time. So we're finding Q prime of 0.5 by substituting 0.5 in for tea, and we end up with 4.75 and the units on charge or cool ums and the time is in seconds. So we have cool arms per second and then for part B. We go through that process again. The charge of time one substituted one into the derivative, and that one is five cool ums per second. And the final question is, when is the current the lowest? So if you think about this current function, recognize that it's a parabola that opens up, and so the lowest would be at the Vertex so we can calculate the Vertex using the formula we learned way back in algebra to the Vertex is the opposite of the over to a comma. Whatever the Y value is now, we only need the X value because we only need the time. So using the numbers from this equation three is a negative. Four is B and six to see we have opposite of Be Over to A is equal to 4/6, and that reduces to 2/3 so the current is the lowest at 2/3 of a second.

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