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Rate of Change of Quantities




Rate of Change of Quantities

  • The derivative of a function at a point represents the rate of change of the function at that point.
  • If y= f(x),Then $\frac {dy}{dx}$ represent the rate of change of y with respect to x
  • Also $\frac {dy}{dx} _{x=x_0}$ represent the rate of change of y with respect to x at $x=x_0$
  • Also y = f(t) and x = f(t), then rate of change of y with respect to x
    $\frac {dy}{dx} = \frac {dy}{dt} / \frac {dx}{dt}$ if $\frac {dx}{dt} \ne 0$

Important Points to remember

$\frac {dy}{dx}$ is positive if y increases as x increases and is negative if y decreases as x increases

Solved Examples

Question 1
A ladder 10 meters long is leaning against a wall. If the bottom of the ladder is being pulled away from the wall at a rate of 2 meters per minute, how fast is the top of the ladder sliding down the wall when the bottom is 6 meters from the wall?
Solution
Using the Pythagoras theorem for the ladder, wall, and ground:
\[ x^2 + y^2 = 100 \]
Differentiate with respect to time \( t \):
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
We are given following as per the Question
\( \frac{dx}{dt} = 2 \) m/min, \( x = 6 \) m
Using Pythagoras theorem:
\[ y = \sqrt{100 - 36} = 8 \]
Plugging values into the differentiated equation:
\[ 2(6)(2) + 2(8) \frac{dy}{dt} = 0 \]
\[ \frac{dy}{dt} = -\frac{12}{8} = -1.5 \]
Therefore
The top of the ladder is sliding down the wall at a rate of 1.5 m/min.

Question 2
A square's side increases at a rate of 3 cm/min. How fast is the area of the square increasing when the side is 10 cm long?
Solution
For a square, \( A = x^2 \).
\[ \frac{dA}{dx} = 2x \]
Given: \( \frac{dx}{dt} = 3 \) cm/min, \( x = 10 \) cm
Therefore
\[ \frac{dA}{dt} = \frac{dA}{dx} \cdot \frac{dx}{dt} = 2(10)(3) = 60 \]
Thus The area is increasing at a rate of $60 /; cm^2/min$.


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