The voltage in a circuit is given by the formula V = 16t3 - 79t. Find the values for t for which V is a maximum or a minimum.
larger t in seconds =
smaller t in seconds =
My Answer:
V=16t^3-79t
V'=48t^2-79
0=48t^2-79
Sq root of 1.645833 = t
1.2829= t
V''=96t positive therefore a min
Larger t = 1.2829
smaller t = -1.2829
Can you please confirm this is correct.
looks good to me, but I wonder whether negative values of t make sense ...
Thanks Steve.
Your calculations are not correct. Let me guide you through the correct steps to find the values of t for which V is a maximum or minimum.
Given the equation V = 16t^3 - 79t, we need to find the critical points by setting the derivative V' equal to zero.
Step 1: Find the derivative of V with respect to t:
V' = dV/dt = 48t^2 - 79
Step 2: Set V' equal to zero and solve for t:
48t^2 - 79 = 0
Step 3: Solve the quadratic equation:
48t^2 = 79
t^2 = 79/48
t = ±sqrt(79/48)
Step 4: Simplify the square root and calculate the approximate values of t:
t ≈ ± 1.1087
Therefore, the approximate values for t when V is a maximum or minimum are t ≈ 1.1087 and t ≈ -1.1087.
To summarize:
Larger t in seconds ≈ 1.1087
Smaller t in seconds ≈ -1.1087
Please note that the values provided are approximate, as we rounded the square root.