Rate of Change of Quantities

Identify the changing quantities; write the formula connecting them; differentiate with respect to the required variable; substitute the given value at the end; attach correct units and direction such as increasing or decreasing.

Usually no diagram is necessary, but a labelled sketch of the shape can help identify the variables in area, volume, or distance questions.

Using dV/dr as the final answer when the question asks for dV/dt.

If y is a function of x, then dy/dx gives the instantaneous rate of change of y with respect to x at a given value of x.

If y is a function of x, then dy/dx gives the instantaneous rate of change of y with respect to x at a given value of x. For y = f(x), instantaneous rate = dy/dx. If y = f(t), rate with respect to time = dy/dt. Chain rule condition: if y depends on x and x depends on t, then dy/dt = (dy/dx)(dx/dt). If the radius r of a circle increases at 2 cm/s, then the area A = pi r^2 changes at dA/dt = 2 pi r dr/dt. At r = 5 cm, dA/dt = 2 pi(5)(2) = 20 pi cm^2/s.

In rate-of-change questions, the derivative must be read in the context of the variables. If y = f(x), then dy/dx measures how many units y changes for one unit change in x at that instant. In related-rate problems, two or more quantities usually depend on time, so each derivative must be taken with respect to t even if the formula first contains radius, side, height, or distance. The main Class 12 skill is to keep the relation variable-based until differentiation is complete, then substitute the instant mentioned in the question. A positive rate means the quantity is increasing, a negative rate means it is decreasing, and the unit of the derivative must combine the units of the two quantities, such as cm^2/s or m/s. For y = f(x), instantaneous rate = dy/dx. If y = f(t), rate with respect to time = dy/dt. Chain rule condition: if y depends on x and x depends on t, then dy/dt = (dy/dx)(dx/dt). A spherical balloon has volume V = (4/3)pi r^3. Its radius increases at 0.5 cm/s. Find dV/dt when r = 6 cm. Differentiate with respect to t: dV/dt = 4 pi r^2 dr/dt. Substitute r = 6 and dr/dt = 0.5: dV/dt = 4 pi(36)(0.5) = 72 pi cm^3/s. The volume is increasing at 72 pi cm^3/s. Usually appears as a short-answer or case-based problem involving area, volume, distance, or height changing with time. Using dV/dr as the final answer when the question asks for dV/dt.