The data entry speed (in entries per minute) of a data clerk trainee is represented by the following equation, where x is the number of hours of training he has had.
S = 10
0.8x + 4
, 0 ≤ x ≤ 100
What is the rate at which his speed is changing and what does this rate mean when he has had the following hours of training?
(a) 15 hours of training
(b) 40 hours of training
Oh, the data entry speed of our trainee, huh? Let's crunch some numbers and sprinkle some humor on top!
(a) When the trainee has had 15 hours of training:
To find the rate at which his speed is changing, we need to take the derivative of the equation with respect to x. So, let's do some math:
S = 0.8x + 4
dS/dx = 0.8
Therefore, the rate at which his speed is changing when he has had 15 hours of training is a whopping 0.8 entries per minute. That's like the speed of light in the world of data entry!
(b) Alright, now let's see what happens when he has had 40 hours of training:
Using the same process, we'll take the derivative and see what we get:
dS/dx = 0.8
Surprise! It turns out that the rate at which his speed is changing is still the same, 0.8 entries per minute. Looks like our trainee has hit a steady pace, not much room for improvement. But hey, at least he's consistent!
So, to sum it up, when our trainee has had 15 hours of training, his speed is changing at a rate of 0.8 entries per minute, and when he has had 40 hours of training, his speed is still changing at the same rate of 0.8 entries per minute. Keep that peppy pace up, trainee!
To find the rate at which the data entry speed is changing, we need to find the derivative of the function representing his speed with respect to time (x).
The given equation for the data entry speed is:
S = 0.8x + 4, 0 ≤ x ≤ 100
Let's find the derivative of the speed function:
dS/dx = d/dx (0.8x + 4)
= 0.8
Therefore, the rate at which his speed is changing is 0.8 entries per minute.
Now, let's calculate the rate of change at the specified hours of training:
(a) When he has had 15 hours of training:
To find the rate of change at 15 hours, substitute 15 for x in the derivative:
dS/dx = 0.8
So, the rate at which his speed is changing at 15 hours of training is 0.8 entries per minute.
(b) When he has had 40 hours of training:
To find the rate of change at 40 hours, substitute 40 for x in the derivative:
dS/dx = 0.8
So, the rate at which his speed is changing at 40 hours of training is also 0.8 entries per minute.
To find the rate at which the data entry speed is changing, we need to calculate the derivative of the given equation with respect to x.
The given equation is: S = 0.8x + 4
Taking the derivative of S with respect to x:
dS/dx = d/dx(0.8x + 4)
The derivative of a constant term (4) is zero, so we only need to differentiate the 0.8x term:
dS/dx = 0.8
Therefore, the rate at which the data entry speed is changing is constant and equal to 0.8.
Now, let's calculate the rate at two different points:
(a) When the trainee has had 15 hours of training:
Substitute x = 15 into the derivative equation:
Rate at 15 hours = dS/dx at x = 15 = 0.8
So, when the trainee has had 15 hours of training, the rate at which his speed is changing is 0.8 entries per minute.
(b) When the trainee has had 40 hours of training:
Substitute x = 40 into the derivative equation:
Rate at 40 hours = dS/dx at x = 40 = 0.8
So, when the trainee has had 40 hours of training, the rate at which his speed is changing is still 0.8 entries per minute.
Therefore, for both 15 and 40 hours of training, the rate at which the data entry speed is changing remains constant at 0.8 entries per minute.