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
The equation for the position of an object at time t is represented by the equation f left parenthesis t right parenthesis equals 4 t squared minus 2 t. Which equation represents the instantaneous velocity at any given time, t?
(1 point)
Responses
v left parenthesis t right parenthesis equals 8 t
Image with alt text: v left parenthesis t right parenthesis equals 8 t
v left parenthesis t right parenthesis equals 4 t
Image with alt text: v left parenthesis t right parenthesis equals 4 t
v left parenthesis t right parenthesis equals 4 t minus 2
Image with alt text: v left parenthesis t right parenthesis equals 4 t minus 2
v left parenthesis t right parenthesis equals 8 t minus 2
(1 point)
Responses
negative 5
Image with alt text: negative 5
0
0
2
2
8
The equation for the position of an object at time t is represented by the equation f left parenthesis t right parenthesis equals 4 t squared minus 2 t. Which equation represents the instantaneous velocity at any given time, t?
(1 point)
Responses
v left parenthesis t right parenthesis equals 8 t
Image with alt text: v left parenthesis t right parenthesis equals 8 t
v left parenthesis t right parenthesis equals 4 t
Image with alt text: v left parenthesis t right parenthesis equals 4 t
v left parenthesis t right parenthesis equals 4 t minus 2
Image with alt text: v left parenthesis t right parenthesis equals 4 t minus 2
v left parenthesis t right parenthesis equals 8 t minus 2
Answers
GPT-4o mini
To find the current at \( t = 0 \) given the charge function \( q(t) = 3t^2 + 2t - 5 \), we first need to compute the current \( I(t) \), which is the derivative of \( q(t) \) with respect to time \( t \).
1. **Calculate the derivative \( I(t) \):**
\[
I(t) = \frac{dq}{dt} = \frac{d}{dt}(3t^2 + 2t - 5)
\]
Using the power rule for differentiation:
\[
I(t) = 6t + 2
\]
2. **Substitute \( t = 0 \) to find the current:**
\[
I(0) = 6(0) + 2 = 2
\]
So, the current at \( t = 0 \) is \( 2 \).
For the second part, the instantaneous velocity \( v(t) \) of an object is the derivative of the position function \( f(t) \) given by \( f(t) = 4t^2 - 2t \).
1. **Calculate the derivative \( v(t) \):**
\[
v(t) = \frac{df}{dt} = \frac{d}{dt}(4t^2 - 2t)
\]
Again using the power rule for differentiation:
\[
v(t) = 8t - 2
\]
Therefore, the equation that represents the instantaneous velocity at any given time \( t \) is:
- \( v(t) = 8t - 2 \)
### Summary of Answers:
1. Current at \( t = 0 \): **2**
2. Instantaneous velocity function: **\( v(t) = 8t - 2 \)**
1. **Calculate the derivative \( I(t) \):**
\[
I(t) = \frac{dq}{dt} = \frac{d}{dt}(3t^2 + 2t - 5)
\]
Using the power rule for differentiation:
\[
I(t) = 6t + 2
\]
2. **Substitute \( t = 0 \) to find the current:**
\[
I(0) = 6(0) + 2 = 2
\]
So, the current at \( t = 0 \) is \( 2 \).
For the second part, the instantaneous velocity \( v(t) \) of an object is the derivative of the position function \( f(t) \) given by \( f(t) = 4t^2 - 2t \).
1. **Calculate the derivative \( v(t) \):**
\[
v(t) = \frac{df}{dt} = \frac{d}{dt}(4t^2 - 2t)
\]
Again using the power rule for differentiation:
\[
v(t) = 8t - 2
\]
Therefore, the equation that represents the instantaneous velocity at any given time \( t \) is:
- \( v(t) = 8t - 2 \)
### Summary of Answers:
1. Current at \( t = 0 \): **2**
2. Instantaneous velocity function: **\( v(t) = 8t - 2 \)**