*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **eters Let this be
length of
his shadow,
Say y
meters W WL represents the distance of
20 meters. L M
Let the
distance of
the man
from the
spotlight be
x meters dx
dy
when x = 5, that is,
= 2.5m / s and we need to find the value of
dt
dt
when the man is 15 meters from the building. We can use similar triangles
x 20
to relate x and y as follows:
=
Þ xy = 40. (When x is 5, y is 8).
2y
dy
dx
Differentiating both sides, we get: x + y
=0
dt
dt
dy
dy -(8)(2.5)
dy
Þ 5 + 8(2.5) = 0 Þ
=
Þ
= -4
dt
dt
5
dt
Hence the man’s shadow is decreasing at 4 m/s when he is 15 meters from the building. 5 Q11. For what values of the constants a and b is (1,6) a point of inflection of the curve
y = x 3 + ax 2 + bx + 1?
Solution
f '( x) = y ' = 3 x 2 + 2ax + b.
f ''( x) = 6 x + 2a.
Since (1,6) is a point of inflection, f ''(1) = 0
Þ 6(1) + 2a = 0 Þ 2a = -6 Þ a = -3
Since (1,6) is a point on the graph, it satisfies the equation of the graph.
Hence, f (1) = 6 Þ 13 + a ×12 + b ×1 + 1 = 6
Þ a+b = 4
\ b = 4 - a = 4 - (-3) =...

View
Full
Document