Q5. For s = 4.9t2, find the average velocity between t = 1.8 and t = 2 (NCERT Table 12.2).
s(1.8) = 4.9 × 3.24 = 15.876, so (19.6 − 15.876)/0.2 = 18.62 m/s.
Q6. Find the average velocity of the falling body between t = 0.5 and t = 2 in two ways: from s(0.5), and from the line v = 4.9(t + 2).
s(0.5) = 4.9 × 0.25 = 1.225, so (19.6 − 1.225)/1.5 = 12.25 m/s. The line gives 4.9 × 2.5 = 12.25 m/s, the same.
Q7. Find the velocity of the falling body at t = 4 seconds.
(4.9(4 + h)2 − 78.4)/h = 39.2 + 4.9h, which approaches 39.2. The velocity is 39.2 m/s.
Q8. Show that the velocity of the falling body at time a is 9.8a m/s.
(4.9(a + h)2 − 4.9a2)/h = 4.9(2ah + h2)/h = 9.8a + 4.9h, which approaches 9.8a.
Q9. Find the slope of the tangent to y = x3 at x = 1.
((1 + h)3 − 1)/h = (3h + 3h2 + h3)/h
= 3 + 3h + h2, which approaches 3.
Q10. Find the rate of change of f(x) = 1/x at x = 2.
(1/(2 + h) − 1/2)/h = −h/(2(2 + h)h)
= −1/(2(2 + h)), which approaches −1/4.
Q11. How fast does the area of a circle change with its radius r when r = 3?
(π(3 + h)2 − 9π)/h = π(6 + h), which approaches 6π: the circumference of the circle.
Q12. Find the slope of the tangent to the line y = 3x + 2 at any point x = a.
(3(a + h) + 2 − 3a − 2)/h = 3 for every h ≠ 0, so the slope is 3 everywhere: a line is its own tangent.