Q.The volume of a sphere is increasing in volume at the rate of 3pi,textcm3/textsec. The rate of change of its radius when radius is dfrac12 cm
Concept understanding — Meaning of Derivative and Rate of Change
The derivative f′(x) of a function carries two equivalent readings, and this chapter leans on both throughout.
As a slope. For the curve y=f(x), the slope of the chord joining (x,f(x)) and (x+h,f(x+h)) is the Newton quotient
hf(x+h)−f(x).
Taking h→0 gives the slope of the curve at (x,f(x)) itself:
f′(x)=limh→0hf(x+h)−f(x).
If θ is the angle the tangent makes with the positive x-axis (measured anticlockwise), then f′(x)=tanθ.
As a rate of change. f′(x)=dxdy is also the instantaneous rate of change of y with respect to x; over an interval [a,b] the average rate of change is the ordinary difference quotient b−af(b)−f(a) (a chord slope), while the derivative at a single point is the instantaneous rate.
Motion along a line. If s=f(t) is the position of an object at time t (measured from a fixed origin, positive direction = forward):
v(t)=dtds,a(t)=dtdv=dt2d2s.
- Speed =∣v(t)∣=dtds — always non-negative, regardless of direction.
- v(t)=0: the particle is momentarily at rest.
- v(t)>0: moving forward; v(t)<0: moving backward.
- The particle changes direction exactly where v(t) changes sign (not merely where it is zero — the sign must flip on either side).
- If the particle reverses direction at time tc∈(t1,t2), the total distance travelled from t1 to t2 is ∣s(tc)−s(t1)∣+∣s(t2)−s(tc)∣ — NOT simply ∣s(t2)−s(t1)∣, since backtracking would otherwise cancel out.
- Near Earth's surface a freely falling body has constant acceleration g≈9.8 m/s2 (32 ft/s2), giving a=−g, v=−gt+v0, s=−21gt2+v0t+s0.
Related rates. A related-rates problem links two or more time-varying quantities through a single equation (areas, volumes, distances, …); differentiating that equation with respect to time t (using the chain rule on every quantity that depends on t) produces an equation relating the rates dtd(⋅), which is then solved for the unknown rate at the instant specified. The recurring workflow is: (1) write the governing geometric/physical relation, (2) differentiate both sides with respect to t, (3) substitute the known values (and any values found from the constraint at that instant), (4) solve for the required rate.
In a related-rates problem, only differentiate AFTER writing the general relation — substituting the specific numeric values too early (before differentiating) silently drops the terms whose rates you actually need.
V=34πr3⇒dtdV=4πr2dtdr; substitute r=21, dtdV=3π.
(1) 3 cm/s
Differentiate the volume formula and substitute the given rate and radius.
Step 1. Differentiate. V=34πr3⇒dtdV=4πr2dtdr.
Step 2. Substitute r=21, dtdV=3π.
3π=4π(41)dtdr=πdtdr ⇒ dtdr=3.
(1) 3 cm/s
- CBSE 2020Set ANNUAL1 markMCQQ.The position of a particle moving along a horizontal line of any time t is given by s(t)=3t2−2t−8. The time at which the particle is at rest, is :(a) t=3(b) t=0(c) t=31(d) t=1
›Reveal solutionSolution
Setting the velocity v(t)=dtds to zero gives the time at which the particle is at rest, t=31.
- The position function is s(t)=3t2−2t−8.
- The velocity is the first derivative of position with respect to time: v(t)=dtds=6t−2.
- The particle is "at rest" precisely when its velocity is zero: v(t)=0.
- Solve 6t−2=0: 6t=2, so t=62=31.
✓Final answerThe particle is at rest at t=31 — option (c).
- CBSE 2019Set ANNUAL1 markMCQQ.The surface area of a sphere when the volume is increasing at the same rate as its radius, is :(a) 4π(b) 34π(c) 1(d) 2π1
›Reveal solutionSolution
When the rate of change of volume equals the rate of change of radius, the sphere's surface area equals 1.
- Volume of a sphere: V=34πr3.
- Differentiate with respect to time t: dtdV=4πr2dtdr.
- We are given dtdV=dtdr (the volume increases at the same rate as the radius).
- Substitute: dtdr=4πr2dtdr.
- Since dtdr=0 in general, divide both sides by dtdr: 1=4πr2.
- But 4πr2 is exactly the formula for the surface area S of the sphere.
- Hence S=4πr2=1.
✓Final answerThe surface area of the sphere is 1 — option (c).
- CBSE 2017Set ANNUAL1 markMCQQ.The distance - time relationship of a moving body is given by y=F(t) then the acceleration of the body is the :(a) Gradient of the velocity/time graph(b) Gradient of the distance/time graph(c) Gradient of the acceleration/time graph(d) Gradient of the velocity/distance graph
›Reveal solutionSolution
Since velocity is dy/dt and acceleration is dv/dt, acceleration is the gradient of the velocity–time graph.
- Given y=F(t) is displacement (distance) as a function of time t.
- Velocity v=dtdy is, by definition, the gradient (slope) of the distance–time graph.
- Acceleration a=dtdv is the rate of change of velocity with respect to time — i.e., the gradient of the velocity–time graph.
- Option (b) actually describes velocity, not acceleration; options (c) and (d) do not correspond to any standard kinematic definition.
✓Final answerAcceleration is the gradient of the velocity/time graph — option (a).
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