Question
The current flowing through an electric circuit is the derivative of the charge as a function of time. If the charge q is given by the equation q left parenthesis t right parenthesis equals 3 t squared plus 2 t minus 5, what is the current at t = 0?
(1 point)
Responses
negative 5
Image with alt text: negative 5
0
0
2
2
8
(1 point)
Responses
negative 5
Image with alt text: negative 5
0
0
2
2
8
Answers
GPT-4o mini
To find the current flowing through the circuit at \( t = 0 \), we first need to find the expression for current \( I(t) \). The current is the derivative of charge \( q(t) \) with respect to time \( t \).
Given:
\[
q(t) = 3t^2 + 2t - 5
\]
Taking the derivative:
\[
I(t) = \frac{dq}{dt} = \frac{d}{dt}(3t^2 + 2t - 5)
\]
Using the power rule, we differentiate term by term:
- The derivative of \( 3t^2 \) is \( 6t \).
- The derivative of \( 2t \) is \( 2 \).
- The derivative of the constant \( -5 \) is \( 0 \).
Thus:
\[
I(t) = 6t + 2
\]
Now, we need to evaluate \( I(t) \) at \( t = 0 \):
\[
I(0) = 6(0) + 2 = 0 + 2 = 2
\]
Therefore, the current at \( t = 0 \) is \( 2 \).
The correct answer is:
**2**
Given:
\[
q(t) = 3t^2 + 2t - 5
\]
Taking the derivative:
\[
I(t) = \frac{dq}{dt} = \frac{d}{dt}(3t^2 + 2t - 5)
\]
Using the power rule, we differentiate term by term:
- The derivative of \( 3t^2 \) is \( 6t \).
- The derivative of \( 2t \) is \( 2 \).
- The derivative of the constant \( -5 \) is \( 0 \).
Thus:
\[
I(t) = 6t + 2
\]
Now, we need to evaluate \( I(t) \) at \( t = 0 \):
\[
I(0) = 6(0) + 2 = 0 + 2 = 2
\]
Therefore, the current at \( t = 0 \) is \( 2 \).
The correct answer is:
**2**