An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Find the derivative of the height function: h ′ ( t ) = 30 − 6 t .
Set the derivative to zero to find the time at maximum height: 30 − 6 t = 0 , which gives t = 5 .
Calculate the maximum height by plugging t = 5 into the original equation: h ( 5 ) = 30 ( 5 ) − 3 ( 5 2 ) = 75 .
The maximum height is 75, and it occurs at t = 5 seconds. 75
Explanation
Understanding the Problem We are given a table with values for time T and height h , where h is a function of time, specifically h ( t ) = 30 t − 3 t 2 . The table provides values of h ( t ) for integer values of t from 0 to 10. The goal is to find the maximum height and the time at which it occurs.
Analyzing the Height Function To find the maximum height, we can analyze the function h ( t ) = 30 t − 3 t 2 . This is a quadratic function, and its graph is a parabola opening downwards. The maximum height occurs at the vertex of the parabola.
Finding the Derivative We can find the vertex by finding the time t at which the derivative of h ( t ) is zero. First, we find the derivative of h ( t ) with respect to t :
h ′ ( t ) = d t d ( 30 t − 3 t 2 ) = 30 − 6 t
Finding the Critical Point Now, we set the derivative equal to zero and solve for t :
30 − 6 t = 0 6 t = 30 t = 6 30 = 5
Verifying Maximum To confirm that this is a maximum, we can find the second derivative of h ( t ) :
h ′′ ( t ) = d t 2 d 2 ( 30 t − 3 t 2 ) = d t d ( 30 − 6 t ) = − 6 Since h ′′ ( t ) = − 6 < 0 , the critical point at t = 5 is indeed a maximum.
Calculating Maximum Height Now we calculate the maximum height by plugging t = 5 into the original equation: h ( 5 ) = 30 ( 5 ) − 3 ( 5 2 ) = 150 − 3 ( 25 ) = 150 − 75 = 75
Final Answer Therefore, the maximum height is 75, and it occurs at t = 5 seconds.
Examples
Understanding projectile motion is crucial in fields like sports and engineering. For example, when designing a catapult or analyzing a baseball trajectory, knowing the maximum height and the time it takes to reach that height is essential. This problem demonstrates how to find these key parameters using calculus, allowing engineers to optimize designs and athletes to improve performance.
