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Instantaneous Rate of Change

Rate of change is a very important concept in mathematics. It is defined as the change in something per unit time. Mainly, there are two types of rate of changes -
(1) Average Rate of Change
(2) Instantaneous Rate of Change

Let us first have a look at an average rate of change. The average rate of change in y with respect to x over an interval [a , b] is given by - $\frac{f(b)-f(a)}{b-a}$

On the other hand, the instantaneous rate of change in y at a point x = a is given below -

$lim_{b \rightarrow a}$$\frac{f(b)-f(a)}{b-a}$

At a particular point, the rate of change is called as an instantaneous rate of change, which is same as the derivative values at the same point. If we consider a function, this rate of change at a particular point is the same as the slope of the tangent line at the same point which is the slope of the curve. When the object is traveling along a straight line, the average velocity is the average rate of change of place with respect to time.
Average velocity = $\frac{f(b) - f(a)}{b - a}$
where, a and b are time intervals.

Average rate of change and Instantaneous rate of change are same over short interval of time.

Related Calculators
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Instantaneous Rate of Change Definition

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Instantaneous rate of change is the interpretation of the derivative of a function at a point. The derivative is the limit of the average rate of change between a fixed point and the other point of the curve which is closer to that point(fixed point). If for a function, limit exists, then it is defined as the instantaneous rate of change at that fixed point.
Instantaneous rate of change is the rate of change at that moment. For example, the value of a function at a point $x = a$ is $f(a)$. If there is a small increment in $x$, which is $\delta x$, then the value of the function at $a + \delta x$ is $f(a + \delta x)$. Hence, the change in the value of the function is $f(a + \delta x) - f(a)$. By the term difference of quotient, we mean the change in the value of the function over the change in the value of $x$ which is $\delta x$.

Instantaneous Rate of Change Formula

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The instantaneous rate of change of the function $y = f(x)$, with respect to $x$, when $x = a$ is given by the formula,

$f'(a) = \lim_{\Delta x\rightarrow 0}$$\frac{\Delta y}{\Delta x}$ = $\lim_{x\rightarrow 0}$$\frac{t(a + h) - (t(a)}{h}$ provided the limit exists.

It should be noted that the instantaneous rate of change of $y = f(x)$ with respect to $x$, when $x = a$ is the slope of the tangent at $x = a$.

Graphical Understanding of Average Rate of Change and Instantaneous Rate of Change:

The slope of the secant line PQ is the average rate of change over the intervals defined by the x - coordinates of the two points that are used to determine the secant line. The slope of the tangent at a point on the graph is equivalent to the instantaneous rate of change of a function at this point.

Instantaneous rate of change Examples:


Given below are some of the examples on instantaneous rate of change.

Solved Examples

Question 1: The motion of a ball through from a vertical height of $10$ feet is given by the relation $s(t) = 10 + 12t + 3t^{2}$, where $t$ is the time in sec. Find the instantaneous rate of change when $t = 2$ sec.
Solution:
The formula for instantaneous rate of change is ,
$s'(t) = \lim_{\Delta x\rightarrow 0}$$\frac{s(t + h) - s(t)}{h}$ when $t = 2$.

$s'(2) = \lim_{\Delta x\rightarrow 0}$$\frac{s(2 + h) - s(2)}{h}$

Consider the given function, $s(t) = 10 + 12t + 3t^{2}$

$s(2) = 10 + 12(2) + 3 (2)^{2} = 10 + 24 + 12 = 46$

$s(2 + h) = 10 + 12(2 + h) + 3(2 + h)^{2}$

$= 10 + 24 + 12h + 3(4 + 4h + h^{2}) = 46 + 24h + 3h^{2}$

$\lim_{h \rightarrow 0}$$\frac{s(2 + h) - s(2)}{h}$ = $\lim_{h \rightarrow 0}$$\frac{46 + 24h + 3h^{2} - 46}{h}$ = $\lim_{h\rightarrow 0}$$\frac{h(24 + 3h)}{h}$ = $\lim_{h \rightarrow 0}(24 + 3h) = 24$

Hence, the instantaneous rate of change when $t = 2$ is $24$ ft/sec.

Question 2: The following table shows the temperature of an oven as it heats from room temperature to 300A° F.

Time(min)

0
2 4 6 8 10
Temperature (A° F )
80 120 160 200 260 300

Find the instantaneous rate of change in the temperature at exactly 5 min using the given data.
Solution:
Using the interval $2\leq t \leq 8$,

$\frac{\Delta y}{\Delta x}$ = $\frac{f(8) - f(2)}{8 - 2}$ = $\frac{260 - 120}{6}$ = $\frac{140}{6}$ = $23.3$ A° F / min

Using the interval $4\leq t \leq 6$, [choosing some centered integral around 5 min]

$\frac{\Delta y}{\Delta x}$ = $\frac{f(6) - f(4)}{6 - 4}$ = $\frac{200 - 160}{2}$ = $\frac{40}{2}$ = $20$ A° F / min

As the centered intervals around 5 min get smaller, we can conclude that the average rate of change in the temperature of the oven is closer to 20 A° F / min.
Hence, the instantaneous rate of change at exactly 5 min is 20 A° F / min

Question 3:
  1. On the graph paper, sketch the graph of $f(x) = x^{2}$.
  2. Draw a secant line passing through the points when $x = 2$ and $x = 4$.
  3. Estimate the location of the point tangent to the secant drawn at this point.

Solution:
The co-ordinate of the points at $x = 2$ and $x = 4$ are $(2,2^{2})$ and $(4,4^{2})$ which is $(2,4)$ and $(4,16)$. The slope of the line joining the two points is

$\frac{f(4) - f(2)}{4 - 2}$ = $\frac{16 - 4}{2}$ = $\frac{12}{2}$ =$6$

The average rate of change when $2 \leq x \leq 4$ is $6$.

Instantaneous Rate of Change

We draw the graph of the function $y = x^{2}$, and join the line joining the points $(2, 4)$ and $(4, 16)$.

The tangent parallel to the line PQ passes through the point $(3,9)$. The instantaneous rate of change is the same as the average rate of change between the points P and Q. [slope of the secant PQ].
Hence, the instantaneous rate of change at $x = 3$ is $6$ units.


Finding Instantaneous Rate of Change

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This is found by calculating the average rate of change over the shortest possible interval as the change in the independent variable heads towards zero. An important difference is that an average rate of change is calculated over an interval, whereas an instantaneous rate of change is calculated at a point in time. For this, we can choose a point in the given interval and do the same procedure(as for the average rate of change).

Given below are some of the examples on finding instantaneous rate of change

Solved Examples

Question 1: Find the instantaneous rate of change for the function $f(x) = 4x^{2} + 4x - 3$ at $x = 5$?
Solution:

$f(x) = 4x^{2} + 4x - 3$

Find the first derivative of the given function.

$f’(x) = 2(4x) + 4(1) - 0$

= $8x + 4$

Since, we have to find the instantaneous rate of change at $x = 5$

$f’(5) = 8( 5) + 4$

= $40 + 4$

= $44$

The instantaneous rate of change for function $f(x) = 4x^{2} + 4x - 3$ at $x = 5$ is $44$



Question 2: Find the instantaneous rate of change for function $f(x) = 6x^{3} + 4x^{2} - 3x$ at $x = 5$?
Solution:

$f(x) = 6x^{3} + 4x^{2} - 3x$

Find the first derivative of the given function.

$f’(x) = 6(3x^{2}) + 4(2x) - 3$

= $18x^{2} + 8x - 3$

Since, we have to find the instantaneous rate of change at $x = 5$

$f’(5) = 18(5^{2}) + 4(2 \times 5) - 3$

= $(18 \times 25) + 4(10) - 3$

= $450 + 40 - 3$

= $490 - 3$

= $487$

The instantaneous rate of change for function $f(x) = 6x^{3} + 4x^{2} - 3x$ at $x = 5$ is $487$



Question 3: Find the instantaneous rate of change for function $f(x) = 6x^{3} + 4x^{2} - 3x$ at $x = 1$?
Solution:

$f(x) = 6x^{3} + 4x^{2} - 3x$

Find the first derivative of the given function.

$f’(x) = 6(3x^{2}) + 4(2x) - 3$

= $18x^{2} + 8x - 3$

Since, we have to find the instantaneous rate of change at $x = 5$

$f’(5) = 18(1^{2}) + 4(2 \times 1) - 3$

= $(18 \times 1) + 4(2) - 3$

= $18 + 8 - 3$

= $26 - 3$

= $23$

The instantaneous rate of change for function $f(x) = 6x^{3} + 4x^{2} - 3x$ at $x = 1$ is $23$

Question 4: Let $y = x^{3} - 2$, find the instantaneous rate of change of $y$ with respect to $x$ over the interval [2,4].
Solution:
We know that an instantaneous rate of change is calculated at a point in the given interval, so let the point be $x = 3$.

Given that $f(x) = x^{3} - 2$

So, $f(3) = 27 - 2 = 25$

Put $x = x_{1}$ in this, then

$f(x_1) = (x_1)^{3} - 2$
Hence, the instantaneous rate of change is

$\lim_{\delta x_{1}\rightarrow 3}$ $\frac{\delta y}{\delta x}$ = $\lim_{\delta x_{1}\rightarrow 3}$ $\frac{x_{1}^{3} - 2 - 25}{x_{1} - 3}$

= $\lim_{\delta x_{1}\rightarrow 3}$ $\frac{x_{1}^{3} - 27}{x_{1} - 3}$

= $\lim_{\delta x_{1}\rightarrow 3}$ $\frac{(x_{1} - 3)(x_{1}^{2} + 3x_{1} + 9)}{x_{1} - 3}$
= $\lim_{\delta x_{1}\rightarrow 3}$ $\frac{(x_{1}^{2} + 3x_{1} + 9)}{1}$

= $3^{2} + 3 \times 3 + 9$

= $9 + 9 + 9$ = $27$.

So, the instantaneous rate of change of $y$ is $27$.

Instantaneous Rate of Change of a Function

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Instantaneous rate of change of a function $f$ at the point $a$ is the limit of the average rate of change of $f$ over shorter intervals around $a$. Instantaneous rate of change of a function $f(x)$ at the point $x = a$ is $f'(x)$ i.e. the value of the derivative at point $a$. If we write rate of change, it means instantaneous rate of change.

Example: The volume V of the sphere is $\frac{4}{3}$$ \pi r^{3}$. So, the rate of change of V with respect to $r$ is $\frac{dV}{dr}$ = $4 \pi r^{2}$.

Instantaneous Rate of Change Practice Problems

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Given below are some of the practice problems on instantaneous rate of change.

Practice Problems

Question 1: For the function $f(x) = 8x^{2} - 7$, estimate the instantaneous rate of change for the given values of $x$.
  • $x = -1$
  • $x = 0$
  • $x = 10$

Question 2: Robert purchased a home for 230,000 dollars. Appreciation of the home's value can be modeled by the function, $f(t) = 230,000(1.05)t$, where $f(t)$ is the value of the home and $t$ is the number of years that the family owns the home. Find the instantaneous rate of change in the home's value $t$ the start of the $5^{th}$ year of owning the home.
Question 3: A spherical balloon is inflated. Estimate the rate at which its volume is changing with respect to the radius when the radius measures $18$ cm.
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