More practice questions

  1. Q5. Find the rate of change of the area of a circle with respect to its radius when (a) r = 3 cm; (b) r = 4 cm (NCERT Exercise 6.1, Q1, page 150).

    Show answerHide answer

    dA/dr = 2πr: (a) 6π cm2 per cm; (b) 8π cm2 per cm.

  2. Q6. An edge of a cube increases at 3 cm/s. How fast is the volume increasing when the edge is 10 cm? And a circle’s radius increases at 3 cm/s: how fast is the area increasing when the radius is 10 cm? (NCERT Exercise 6.1, Q4 and Q3, page 150)

    Show answerHide answer

    dV/dt = 3x2 dx/dt = 3(100)(3) = 900 cm3/s. dA/dt = 2πr dr/dt = 2π(10)(3) = 60π cm2/s.

  3. Q7. A balloon stays spherical, with diameter (3/2)(2x + 1). Find the rate of change of its volume with respect to x (NCERT Exercise 6.1, Q13, page 151).

    Show answerHide answer

    A positive diameter requires x > −1/2. The radius is (3/4)(2x + 1), so V = (4/3)π(27/64)(2x + 1)3 = (9π/16)(2x + 1)3. Then dV/dx = (9π/16) × 3(2x + 1)2 × 2 = (27π/8)(2x + 1)2.

  4. Q8. R(x) = 13x2 + 26x + 15. Find the marginal revenue when x = 7 (NCERT Exercise 6.1, Q16, page 151).

    Show answerHide answer

    MR = 26x + 26 = 182 + 26 = ₹208.

  5. Q9. The radius of a circle increases at 0.7 cm/s. What is the rate of increase of its circumference? (NCERT Exercise 6.1, Q6, page 151)

    Show answerHide answer

    C = 2πr, so dC/dt = 2π × 0.7 = 1.4π cm/s, whatever the radius.

  6. Q10. A spherical balloon has a variable radius. Find the rate at which its volume changes with the radius when the radius is 10 cm (NCERT Exercise 6.1, Q9, page 151).

    Show answerHide answer

    dV/dr = 4πr2 = 400π cm3 per cm.

  7. Q11. Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 6 cm.

    Show answerHide answer

    dV/dS = (dV/dr) ÷ (dS/dr) = 4πr2 ÷ 8πr = r/2, so 3 cm3 per cm2 at r = 6 cm. No time is needed: the chain rule works through r, as NCERT’s works through t (page 147).

  8. Q12. A man 1.6 m tall walks away from a lamp 4.8 m high at 1.2 m/s. How fast does his shadow lengthen, and how fast does the tip of the shadow move?

    Show answerHide answer

    With the man x m from the post and a shadow s m long, similar triangles give s/1.6 = (x + s)/4.8, so s = x/2. Then ds/dt = 0.6 m/s, at every distance. The tip is x + s from the post, so it moves at 1.2 + 0.6 = 1.8 m/s: the shadow’s length and its tip change at different rates. NCERT works a version in Miscellaneous Example 32 (pages 178–179).