Question

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instantaneous rate of change of y with respect to x at the value x=3. Average Rate of Change : Instantaneous Rate of Change at x=3:

Answer 1

The instantaneous rate of change of
`y` with respect to
`x` at the value
`x = 3` is 18.

Explanation:

a) Geometrically speaking, the average rate of change of
`y` with respect to
`x` over the interval by definition of secant line:

`r = (y(b) -y(a))/(b-a)` (1)

Where:

`a`,
`b` - Lower and upper bounds of the interval.

`y(a)`,
`y(b)` - Function exaluated at lower and upper bounds of the interval.

If we know that
`y = 3\cdot x^(2)`,
`a = 3` and
`b = 6`, then the average rate of change of
`y` with respect to
`x` over the interval is:

`r = (3\cdot (6)^(2)-3\cdot (3)^(2))/(6-3)`

`r = 27`

The average rate of change of
`y` with respect to
`x` over the interval
`[3,6]` is 27.

b) The instantaneous rate of change can be determined by the following definition:

`y' = \lim_(h \to 0)(y(x+h)-y(x))/(h)` (2)

Where:

`h` - Change rate.

`y(x)`,
`y(x+h)` - Function evaluated at
`x` and
`x+h`.

If we know that
`x = 3` and
`y = 3\cdot x^(2)`, then the instantaneous rate of change of
`y` with respect to
`x` is:

`y' = \lim_(h \to 0) (3\cdot (x+h)^(2)-3\cdot x^(2))/(h)`

`y' = 3\cdot \lim_(h \to 0) ((x+h)^(2)-x^(2))/(h)`

`y' = 3\cdot \lim_(h \to 0) (2\cdot h\cdot x +h^(2))/(h)`

`y' = 6\cdot \lim_(h \to 0) x +3\cdot \lim_(h \to 0) h`

`y' = 6\cdot x`

`y' = 6\cdot (3)`

`y' = 18`

The instantaneous rate of change of
`y` with respect to
`x` at the value
`x = 3` is 18.