Here you will learn derivative as a rate measure class 12 with examples.
Letโs begin โ
Derivative as a Rate Measure
Whenever one quantity y varies with another quantity x, satisfying some rule y = f(x), then
\(dy\over dx\) (or f'(x)) represents the rate of change of y with respect to x
and \(({dy\over dx})_{x=a}\) (or f'(a)) represents the rate of change of y with respect to x at x = a.
Remark 1 : If x = \(\phi(t)\) and y = \(\psi(t)\) , then \(dy\over dx\) = \(dy/dt\over dx/dt\) , provided that \(dx\over dt\) \(\ne\) 0.
Thus, the rate of change y with respect to x can be calculated by using the rate of change of y and that of x each with respect to t.
Remark 2 : The term โrate of changeโ will mean the instantaneous rate of change unless stated otherwise.
Example : The volume of a cube is increasing at a rate of 9 \(cm^3/s\). How fast is the surface area increasing when the length of an edge is 10cm ?
Solution : Let x be the length of side, V be the volume and S be the surface area of the cube.ย
Then V = \(x^3\) and S = \(6x^2\) , where x is a function of time t.
\(dV\over dt\) = 9 \(cm^3/s\) = \(d\over dt\)(\(x^3\)) = \(3x^2\)\(dx\over dt\)
ย \(dx\over dt\) = \(3\over x^2\)
Now, S = \(6x^2\)
Differentiating it with respect to t,
\(dS\over dt\) = \(d\over dt\)(\(6x^2\))ย
= 12x(\(3\over x^2\)) = \(36\over x\)
\(\implies\) \(({dS\over dt})_{x=10cm}\) = 3.6 \(cm^2/s\)
Example : find the rate of change of the area of the circle with respect to its radius. How fast is the area changing with respect to the radius when the radius is 3cm?
Solution : Let A be the area of the circle. Then.
A = \(\pi r^2\) \(\implies\) \(dA\over dr\) = \(2\pi r\)
Thus, the rate of change of the area of the circle with respect to its radius r is \(2\pi r\).
When r = 3 cm, we obtain
\(dA\over dr\) = \(2\pi \times 3\) cm = \(6\pi\) cm
