Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6cm and a height of 12cm, at the rate of 3 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep? Give an exact answer showing all work and include units in your answer. (10 points)

Therefore the depth of the water is changing at the instant when the water in the tank is 9 cm deep at rate / min.

Step-by-step explanation:

Given that,

Radius of the cone(r)= 6 cm

Height of the cone (h)= 12 cm

The volume of the cone is (V)

Putting

Differentiating with respect to t

....(1)

Given that water is drained out of tank at the rate 3 / min.

It means the rate change of volume is  3 / min that is  

Putting the value of in equation (1)

To find the rate of the depth of water changing at 9 cm depth, we need to put h=9 cm in the above equation.

              / min.

Therefore the depth of the water is changing at the instant when the water in the tank is 9 cm deep at rate / min.

Question:

The options are;

(a) (9 - 2s1, 3 + 3s1, s1)

solution

not a solution

(b) (-4 - 5s1, s1,  -(3 + s1)/2)

solution

not a solution

(c) (11 + 10s1, -3 - 2s1, s1)

solution

not a solution

(d) ((6 - 4s1)/3, s1, -(7 - s1)/4)

solution

not a solution

The options that form a solution of the given system are;

(b) and (c)

Step-by-step explanation:

Here we have

3·x₁ + 8·x₂ − 14·x₃ = 9

x₁ + 3·x₂ − 4·x₃ = 1

(a) (9 - 2·s₁, 3 + 3·s₁,s₁)

3·(9 - 2·s₁) + 8·(3 + 3·s₁) − 14·s₁ = 4s₁+51

Not a solution

(b) (-4 - 5s₁, s₁, -(3 + s1)/2)

3·(-4 - 5s₁) + 8·(s₁) − 14·-(3 + s1)/2 = 9

(-4 - 5s₁) + 3·(s₁) − 4·-(3 + s1)/2 = 2

Solution

(c) (11 + 10s₁, -3 - 2s₁, s₁ )

3·(11 + 10s₁) + 8·(-3 - 2s₁) − 14·s₁  = 9

(11 + 10s₁) + 3·(-3 - 2s₁) − 4·s₁ = 2

Solution

(d) ((6 - 4s1)/3, s1, -(7 - s1)/4)

3·(6 - 4s1)/3+ 8·s1− 14·-(7 - s1)/4 = 0.5s₁ +30.5  

Not a solution