Question

A container in the shape of a hollow cone of semi-vertical angle 45° is held with its vertex pointing downwards. Water is poured into a container at the rate of 6cm³/s.Find the rate at which the depth of the water in the cone is increasing when the depth of the water is 4cm.

Answer 1

The rate at which the depth of the water in the cone is increasing is 0.12 cm/sec.

Step-by-step explanation:

The semi-vertical angle of the hollow cone is 45°.

So, its base radius (r) = height (h)

Now, the volume of a cone is given by, $V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi h^{3}$ {As r = h}

So, differentiating the above equation with respect to time (t in seconds), we get

$\frac{dV}{dt} = \pi h^{2} \frac{dh}{dt}$

⇒ $6 = \frac{22}{7}\times (4)^{2} \times \frac{dh}{dt}$

⇒ $\frac{dh}{dt} = 0.12$ cm/seconds. (Answer)

Answer 2

Answer:

Step-by-step explanation:

The semi-vertical angle of the hollow cone is 45°.

So, its base radius (r) = height (h)

Now, the volume of a cone is given by, V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi h^{3}V=

3

1

πr

2

h=

3

1

πh

3

{As r = h}

So, differentiating the above equation with respect to time (t in seconds), we get

\frac{dV}{dt} = \pi h^{2} \frac{dh}{dt}

dt

dV

=πh

2

dt

dh

⇒ 6 = \frac{22}{7}\times (4)^{2} \times \frac{dh}{dt}6=

7

22

×(4)

2

×

dt

dh

⇒ \frac{dh}{dt} = 0.12

dt

dh

=0.12 cm/seconds. (Answer)