If the volume of spherical ball is increasing at the rate of 4 cm3/s, then the rate of change of its surface area when the volume is 288 cm3, is
The equation of displacement of a particle is x(t) = 5t2 - 7t + 3. The acceleration at the moment when its velocity becomes 5 m/sec is
3 m/sec2
7 m/sec2
10 m/sec2
8 m/sec2
C.
10 m/sec2
We have,
x(t) = 5t2 - 7t + 3 ...(i)
By differentiating both side w.r.t. 't', we get
Velocity ofthe particle,
...(ii)
We have to find acceleration at the moment, when velocity becomes 5 m/sec
From Eq. (ii), we get
10t - 7 = 5
Now, on differentiationg both side of Eq. (11) w.r.t. 't', we get acceleration of the particle
Here, acceleration is constant all the time,
The function
increases in (0, 1) but decreases in (1, 2)
decreases in (0, 2)
increases m (1, 2) but decreases in (0, 1)
increases in (0, 2)
The points of the curve y = x3 + x - 2 at which its tangent are parallel to the straight line y = 4x - 1 are
(2, 7), (- 2, - 11)
(0, 2), (21/3, 21/3)
(- 21/3, - 21/3), (0, - 4)
(1, 0), (- 1, - 4)
The equation of the normal to the curve y = - + 2 at the point of its intersection with the bisector of the first quadrant is
4x - y + 16 = 0
4x - y = 16
2x - y - 1 = 0
2x - y + 1 = 0