Respuesta :
Answer:
a. 10.2 in²/in
10.02 in²/in
b. 10 in²/in
Step-by-step explanation:
The area of a square is given by the following function:
[tex]f(x)=x^2[/tex]
The average rate of change between two points a and b is:
[tex]r = \frac{f(b)-f(a)}{b-a}[/tex]
a. For x = 5 and 5.2:
[tex]r = \frac{5.2^2-5^2}{5.2-5} \\r=10.2\ in^2/in[/tex]
For x = 5 and 5.02
[tex]r = \frac{5.02^2-5^2}{5.02-5} \\r=10.02\ in^2/in[/tex]
b. The instantaneous rate of change in area with respect to x is given by:
[tex]f'(x)=2x[/tex]
For x = 5
[tex]f'(5) =2*5 = 10\ in^2/in[/tex]
Answer:
A) As x changes from 5 inches to 5.2 inches , average rate of change = 10.2π in²/in
- As x changes from 5 inches to 5.02 inches , average rate of change = 10.02π in²/in
B) Instantaneous rate at x=5 is; 10π in²/in
Step-by-step explanation:
A) We are given that the function which is the area of the square is given by:
f(x) = πx²
As x changes from 5 to 5.2, a=5 and h = 0.2
Average rate of change between 2 points A and B is given by;
[f(a+h) - f(a)]/h
Thus, at a=5 and h = 0.2, we have;
[f(5+0.2) - f(5)]/0.2
= [f(5.2) - f(5)]/0.2
Since, f(x) = πx², let's substitute for x to obtain ;
[(π•5.2²) - (π•5²)]/0.2 = 10.2π in²/in
As x changes from 5 to 5.02, a=5 and h = 0.02
Average rate of change between 2 points A and B is given by;
[f(a+h) - f(a)]/h
Thus, at a=5 and h = 0.02, we have;
[f(5+0.02) - f(5)]/0.02
= [f(5.02) - f(5)]/0.02
Since, f(x) = πx², let's substitute for x to obtain ;
[(π•5.02²) - (π•5²)]/0.02 = 10.02π in²/in
B) The instantaneous rate of change of the area with respect to x will be the derivative of f(x).
Thus, since f(x) = πx²
f'(x) = 2πx
So, at x = 5 inches,
Instantaneous rate = 2π(5) = 10π
