I have three questions I need help solving
1) A ball's position, in meters, as it travels every second is represented by the position function s(t) = 4.9t2 + 550.
What is the velocity of the ball after 3 seconds?
2) The cost in dollars of producing x units of a particular telephone is C(x) = x^2 - 2500.
- Find the average rate of change of C with respect to x when the production level is changed from x = 100 to x = 103. Include units in your answer.
- Find the instantaneous rate of change of C with respect to x when x = 100. Include units in your answer.
3) Find lim --> 4- (|x-4|)/(x-4)
(limit approaches from the left hence the negative sign)
Thank you in advance
1) To find the velocity of the ball after 3 seconds, we need to take the derivative of the position function s(t). The derivative gives us the rate of change of the position with respect to time, which is the velocity.
The position function is given by s(t) = 4.9t^2 + 550.
To find the velocity, we take the derivative of s(t) with respect to t:
v(t) = d/dt (4.9t^2 + 550).
Taking the derivative of each term separately, we get:
v(t) = 9.8t.
Now we substitute t = 3 seconds into the velocity function:
v(3) = 9.8 * 3 = 29.4 meters per second.
Therefore, the velocity of the ball after 3 seconds is 29.4 meters per second.
2) a) To find the average rate of change of C with respect to x when the production level is changed from x = 100 to x = 103, we need to find the change in C and divide it by the change in x.
The cost function is given by C(x) = x^2 - 2500.
Let's first find the change in C:
ΔC = C(103) - C(100).
Substitute the values of x into the cost function and evaluate:
ΔC = (103^2 - 2500) - (100^2 - 2500).
Now we can find the change in x:
Δx = 103 - 100.
Finally, we can calculate the average rate of change:
Average rate of change of C with respect to x = ΔC/Δx.
b) To find the instantaneous rate of change of C with respect to x when x = 100, we need to take the derivative of the cost function C(x) and evaluate it at x = 100.
The cost function is given by C(x) = x^2 - 2500.
Taking the derivative of C(x) with respect to x, we get:
C'(x) = 2x.
Now substitute x = 100 into the derivative to find the instantaneous rate of change:
Instantaneous rate of change of C with respect to x = C'(100) = 2 * 100.
Therefore, the average rate of change of C with respect to x when the production level is changed from x = 100 to x = 103 is (ΔC/Δx), and the instantaneous rate of change of C with respect to x when x = 100 is (C'(100)).
3) To find the limit as x approaches 4 from the left of the expression |x - 4| / (x - 4), we substitute x = 4 into the expression:
lim (x→4-) (|x-4|)/(x-4) = (|4-4|)/(4-4).
Since |4-4| = 0 and 4-4 = 0 as well, the expression becomes:
lim (x→4-) (0)/(0).
This is an indeterminate form, as both the numerator and denominator approach zero.
To further evaluate the limit, we can use the fact that |x - 4| = -(x - 4) for x < 4.
Using this, the expression becomes:
lim (x→4-) (-(x-4))/(x-4).
Simplifying, the expression becomes:
lim (x→4-) -1.
Therefore, the limit as x approaches 4 from the left of |x - 4| / (x - 4) is -1.
1. velocity is s', or ds/dt=9.8t
2. dC/dx=2x
average rate of change= (c(103)-c(100))/3
3.
from the - side, lim=-( abs(-e)/-e=1