Round 38562.1 to 1 significant figure

Water is being pumped into a conical tank that is 8 feet tall and has a diameter of 10 feet. If the water is being pumped in at

Deffense [45]

The rate of change of the depth of water in the tank when the tank is half

filled can be found using chain rule of differentiation.

When the tank is half filled, the depth of the water is changing at 1.213 ×

10⁻² ft.³/hour.

Reasons:

The given parameter are;

Height of the conical tank, h = 8 feet

Diameter of the conical tank, d = 10 feet

Rate at which water is being pumped into the tank, = 3/5 ft.³/hr.

Required:

The rate at which the depth of the water in the tank is changing when the

tank is half full.

Solution:

The radius of the tank, r = d ÷ 2

∴ r = 10 ft. ÷ 2 = 5 ft.

Using similar triangles, we have;

$\dfrac{r}{h} = \dfrac{5}{8}$

The volume of the tank is therefore;

$V = \mathbf{\dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h}$

$r = \dfrac{5}{8} \times h$

Therefore;

$V = \dfrac{1}{3} \cdot \pi \cdot \left( \dfrac{5}{8} \times h\right)^2 \cdot h = \dfrac{25 \cdot h^3 \cdot \pi}{192}$

By chain rule of differentiation, we have;

$\dfrac{dV}{dt} = \mathbf{\dfrac{dV}{dh} \cdot \dfrac{dh}{dt}}$

$\dfrac{dV}{dh}=\dfrac{d}{h} \left( \dfrac{25 \cdot h^3 \cdot \pi}{192} \right) = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64}}$

$\dfrac{dV}{dt} = \dfrac{3}{5} \ ft.^3/hour$

Which gives;

$\dfrac{3}{5} = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64} \times \dfrac{dh}{dt}}$

When the tank is half filled, we have;

$V_{1/2} = \dfrac{1}{2} \times \dfrac{1}{3} \times \pi \times 5^2 \times 8 =\mathbf{ \dfrac{25 \cdot h^3 \cdot \pi}{ 192}}$

Solving gives;

h³ = 256

h = ∛256

$\dfrac{3}{5} \times \dfrac{64}{25 \cdot h^2 \cdot \pi} = \dfrac{dh}{dt}$

Which gives;

$\dfrac{dh}{dt} = \dfrac{3}{5} \times \dfrac{64}{25 \cdot (\sqrt[3]{256}) ^2 \cdot \pi} \approx \mathbf{1.213\times 10^{-2}}$

When the tank is half filled, the depth of the water is changing at 1.213 × 10⁻² ft.³/hour.

Learn more here:

brainly.com/question/9168560

6 0

3 years ago

"A 12-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from t

Svet_ta [14]

Answer:

Step-by-step explanation:

The question has typographical errors. The correct question is:

"A 12-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 3 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall?

Solution:

The ladder forms a right angle triangle with the ground. The length of the ladder represents the hypotenuse.

Let x represent the distance from the top of the ladder to the ground(opposite side)

Let y represent the distance from the foot of the ladder to the base of the wall(adjacent side)

The bottom of the ladder is sliding along the pavement directly away from the building at 3ft/sec. This means that y is increasing at the rate of 3ft/sec. Therefore,

dy/dt = 3 ft/s

The rate at which x is reducing would be

dx/dt

Applying Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side², it becomes

x² + y² = 12²- - - - - - - -1

Differentiating with respect to time, it becomes

2xdx/dt + 2ydy/dt = 0

2xdx/dt = - 2ydy/dt

Dividing through by 2x, it becomes

dx/dt = - y/x ×dy/dt- - - - - - - - - - 2

Substituting y = 5 into equation 1, it becomes

x² + 5 = 144

x² = 144 - 25 = 119

x = √119 = 10.91

Substituting x = 10.91, dy/dt = 3 and y = 5 into equation 2, it becomes

dx/dt = - 5/10.91 × 3

dx/dt = - 1.37 ft/s

7 0

4 years ago

Write this ratio as a fraction in simplest form without any units. 42 to 2 days

Korvikt [17]

The ratio in it's simplest form is 21: 1

Ratio can be described as the comparison of two numbers together.

It can be calculated as follows

= 42/1

= 21/1

= 21:1

Hence the ratio is 21:1