Find the rate of change of the area of a circle with respect to its radius r when (a) r=3 cm (b) r=4 cm.
Hint. Differentiate A=pi r^2 with respect to r, then substitute each value.
A=pi r^2, so dA/dr=2 pi r. (a) At r=3: dA/dr=6 pi. (b) At r=4: dA/dr=8 pi.
✦ (a) 6 pi cm^2/cm (b) 8 pi cm^2/cm
The volume of a cube is increasing at the rate of 8 cm^3/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Hint. Relate V and S through the edge length x, differentiate both with respect to time, then combine using the given dV/dt.
V=x^3, so dV/dt=3x^2.dx/dt. Given dV/dt=8, so dx/dt=8/(3x^2). S=6x^2, so dS/dt=12x.dx/dt=12x.[8/(3x^2)]=32/x. At x=12: dS/dt=32/12=8/3.
✦ dS/dt = 8/3 cm^2/s
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Hint. Differentiate A=pi r^2 with respect to time using the chain rule.
A=pi r^2, so dA/dt=2 pi r.dr/dt. Given dr/dt=3, at r=10: dA/dt=2 pi(10)(3)=60 pi.
✦ dA/dt = 60 pi cm^2/s
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Hint. Differentiate V=x^3 with respect to time.
V=x^3, so dV/dt=3x^2.dx/dt. Given dx/dt=3, at x=10: dV/dt=3(100)(3)=900.
✦ dV/dt = 900 cm^3/s
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Hint. Differentiate A=pi r^2 with respect to time.
This is the same relationship as the previous question: A=pi r^2, so dA/dt=2 pi r.dr/dt. Given dr/dt=5, at r=8: dA/dt=2 pi(8)(5)=80 pi.
✦ dA/dt = 80 pi cm^2/s
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Hint. Differentiate C=2 pi r with respect to time; note the rate is constant since C is linear in r.
C=2 pi r, so dC/dt=2 pi.dr/dt=2 pi(0.7)=1.4 pi. Since this relationship is linear in r, the rate is the same regardless of the current radius.
✦ dC/dt = 1.4 pi cm/s
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x=8cm and y=6cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Hint. Differentiate P=2(x+y) and A=xy with respect to time, using the product rule for A.
Given dx/dt=-5, dy/dt=4. (a) P=2(x+y), so dP/dt=2(dx/dt+dy/dt)=2(-5+4)=-2. (b) A=xy, so dA/dt=x.dy/dt+y.dx/dt=8(4)+6(-5)=32-30=2.
✦ (a) Perimeter decreasing at 2 cm/min (b) Area increasing at 2 cm^2/min
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Hint. Differentiate V=(4/3)pi r^3 with respect to time, then solve for dr/dt.
V=(4/3)pi r^3, so dV/dt=4 pi r^2.dr/dt. Given dV/dt=900, at r=15: dr/dt=900/(4 pi (225))=900/(900 pi)=1/pi.
✦ dr/dt = 1/pi cm/s
A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is 10 cm.
Hint. This asks for dV/dr directly (not dV/dt), so just differentiate V with respect to r.
V=(4/3)pi r^3, so dV/dr=4 pi r^2. At r=10: dV/dr=4 pi(100)=400 pi.
✦ dV/dr = 400 pi cm^3/cm
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Hint. Set up the Pythagorean relationship x^2+y^2=25 between the base distance x and wall height y, then differentiate implicitly with respect to time.
x^2+y^2=25 (ladder length constant). Differentiating: 2x.dx/dt+2y.dy/dt=0. At x=4, y=sqrt(25-16)=3, with dx/dt=2: 2(4)(2)+2(3).dy/dt=0, so dy/dt=-16/6=-8/3.
✦ The height is decreasing at 8/3 cm/s
A particle moves along the curve 6y=x^3+2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Hint. Differentiate the curve's equation implicitly with respect to time, then set dy/dt=8.dx/dt and solve for x.
Differentiating 6y=x^3+2 with respect to t: 6.dy/dt=3x^2.dx/dt. Given dy/dt=8.dx/dt: 6(8.dx/dt)=3x^2.dx/dt, so 48=3x^2 (dividing by dx/dt, nonzero), giving x^2=16, so x=4 or x=-4. Substituting back into the curve equation: at x=4, y=(64+2)/6=11. At x=-4, y=(-64+2)/6=-31/3.
✦ The points are (4, 11) and (-4, -31/3)
The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Hint. Differentiate V=(4/3)pi r^3 with respect to time.
V=(4/3)pi r^3, so dV/dt=4 pi r^2.dr/dt. At r=1, dr/dt=1/2: dV/dt=4 pi(1)(1/2)=2 pi.
✦ dV/dt = 2 pi cm^3/s
A balloon, which always remains spherical, has a variable diameter (3/2)(2x+1). Find the rate of change of its volume with respect to x.
Hint. Write the radius in terms of x, substitute into the volume formula, then differentiate with respect to x directly.
Radius r=(3/4)(2x+1). V=(4/3)pi r^3=(4/3)pi[(3/4)(2x+1)]^3. Differentiating with respect to x using the chain rule gives dV/dx=(27 pi/8)(2x+1)^2.
✦ dV/dx = (27.pi/8).(2x+1)^2
Sand is pouring from a pipe at the rate of 12 cm^3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Hint. Use the given ratio to write the cone's volume in terms of height alone, then differentiate with respect to time.
Since h=r/6, r=6h. V=(1/3)pi r^2 h=(1/3)pi(36h^2)h=12 pi h^3. Differentiating: dV/dt=36 pi h^2.dh/dt. Given dV/dt=12, at h=4: dh/dt=12/(36 pi (16))=1/(48 pi).
✦ dh/dt = 1/(48.pi) cm/s
The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x)=0.007x^3-0.003x^2+15x+4000. Find the marginal cost when 17 units are produced.
Hint. Marginal cost is dC/dx, evaluated at x=17.
MC=dC/dx=0.021x^2-0.006x+15. At x=17: MC=0.021(289)-0.006(17)+15=6.069-0.102+15=20.967.
✦ Marginal cost is approximately Rs 20.97
The total revenue in Rupees received from the sale of x units of a product is given by R(x)=13x^2+26x+15. Find the marginal revenue when x=7.
Hint. Marginal revenue is dR/dx, evaluated at x=7.
Since marginal revenue is the derivative of the revenue function, MR=dR/dx=26x+26. At x=7: MR=26(7)+26=182+26=208.
✦ Marginal revenue is Rs 208
The rate of change of the area of a circle with respect to its radius r at r=6 cm is (A) 10 pi (B) 12 pi (C) 8 pi (D) 11 pi.
Hint. Differentiate A=pi r^2 and substitute r=6.
Since the rate of change of A with respect to r is just dA/dr, differentiate: dA/dr=2 pi r. At r=6: dA/dr=12 pi.
✦ (B) 12 pi
The total revenue in Rupees received from the sale of x units of a product is given by R(x)=3x^2+36x+5. The marginal revenue, when x=15 is (A) 116 (B) 96 (C) 90 (D) 126.
Hint. Differentiate R(x) and substitute x=15.
This mirrors the marginal revenue calculation from Exercise 6.1 Q16: MR=dR/dx=6x+36, because marginal revenue is the revenue function's derivative. At x=15: MR=6(15)+36=90+36=126.
✦ (D) 126
Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh106.pdf) — Exercise 6.1 (18 questions), Exercise 6.2 (19 questions), Exercise 6.3 (29 questions), plus the chapter's Miscellaneous Exercise (16 questions), 82 questions total. Exercise pages were rendered as 300dpi images to read the piecewise/fractional notation accurately. The old stub taught tangents and normals as a full section with the standard slope formulas, but that topic does not appear anywhere in the current book's four actual sections (6.2 Rate of Change, 6.3 Increasing and Decreasing Functions, 6.4 Maxima and Minima) — confirmed by a full-text search finding only a single mention of 'tangent and normal,' inside the chapter's own introduction paragraph, which also promises 'approximate value of certain quantities' (differentials/approximations) that likewise never appears in any actual section. Both are leftover references from an older, longer edition. The old stub also collapsed the chapter's three real exercises into one invented 40-question group with no solutions file behind it. Every numeric answer in this file (rates of change, monotonicity intervals, local/absolute extrema, and every optimisation word problem including the classic box, cylinder, cone-in-sphere, and cone-in-cone problems) was independently verified with a Python sympy script, and every optimisation problem's extraneous or degenerate critical point (like the box-folding problems' second root that collapses the volume to zero) was explicitly checked and discarded before finalising the answer.. Questions are referenced from the NCERT textbook for identification.
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