Mathematics, 22.11.2020 09:50 nicholasferrell

Water runs into a conical tank at the rate of $9ft^{3}/min.$ The tank stands point down and has a height of 10 ft and a base of 5 ft. How fast is the water level rising when the water is 6 ft deep?

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Answer from: foxesrcool11

$\frac{1}{\pi} \frac{\text{ft}}{\text{min}}$ is the rate at which the water is rising when the water is 6 ft deep.

Step-by-step explanation:

See the attached diagram that I drew to represent this problem.

To solve related rates problems, let's use the recommended steps:

Draw a diagram.Label all quantities and their rates of change.Relate all quantities in the same equation.Differentiate (implicitly) with respect to time.Use the resulting equation to answer the question in context.Step 1:

I already drew the diagram; see attached image.

Step 2:

I labeled the quantities we are given in the problem. h = 10 ft, and r = 5 ft. We are also told that the change in volume is 9 ft³/min; dV/dt = 9 ft³/min.

We want to find dh/dt when h = 6.

$\frac{dh}{dt}\ \vert \ _h_=_6 =\ ?$ Step 3:

We know that we are dealing with a cone in this problem, and we are given the volume of the cone. Therefore, we can use the formula for the volume of a cone in order to relate all of the quantities in the same equation.

$V=\frac{1}{3} \pi r^2 h$

Since we only want the two variables, V and h, we can solve for r in terms of h and substitute this value for r in the formula.

This is because when we perform implicit differentiation, we do not have the change in r (dr/dt) but we do have dh/dt, which is what we are trying to solve for.

We know that r = 5 ft, and h = 10 ft. Therefore, we can say that $\frac{r}{h}=\frac{5}{10} \rightarrow \frac{r}{h} = \frac{1}{2}$ . Multiply h to both sides to solve for the variable r: $r=\frac{h}{2}$ .

Substitute this into the volume of a cone equation:

$V=\frac{1}{3} \pi(\frac{h}{2})^2 h$

Simplify this equation.

$V=\frac{\pi}{3}\cdot \frac{h^2}{4} \cdot h$ $V=\frac{\pi}{12}h^3$ Step 4:

Perform implicit differentiation on the volume equation.

$\frac{dV}{dt} =\frac{\pi}{12}3h^2 \cdot \frac{dh}{dt}$ Step 5:

Substitute known values and solve for dh/dt to find the change in height, or the rise in water level, when the water is 6 ft deep (h = 6).

We know that:

$\frac{dV}{dt} =9$ h=6

Plug these values into the implicitly differentiated volume equation.

$9=\frac{\pi}{12}3(6)^2\cdot \frac{dh}{dt}$ $9=\frac{3\pi}{12}\cdot 36 \cdot \frac{dh}{dt}$ $9=9\pi\frac{dh}{dt}$ $\frac{dh}{dt} =\frac{9}{9\pi}= \frac{1}{\pi}$

Answering the question in context:

Since dh/dt = 1/π when h = 6 ft, we can say that the water is rising at a rate of $\frac{1}{\pi} \frac{\text{ft}}{\text{min}}$ when the water is 6 ft high.

Water runs into a conical tank at the rate of The tank stands point down and has a height of 10 ft

Answer from: nschavez123

Hey there!

We can create some variables to denote some of the things we are working with.

v = volume

h = height

r = radius

We know that [ dv/dt = (π/3)r²h ]. We also know that [ r/h = 5/10 = 1/2 ] which resembles the parts of a triangle.

Solution:

v = π/3*h/2²

h = πh³/(4)(3)

~Differentiate both sides

dv/dt = (3πh³/4*3)(dh/dt)

~Simplify

dh/dt = (4dv/dt)/πh²

~Use chain rule

(4)(9)/π6²

1/π

Thus, the speed of the water level rising is [ 1/π ft/min ] when the water is 6ft deep.

Best of Luck!

Answer from: Quest

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

Answer from: Quest

2x + 3y = 114x + 2y = - 2 ( original equation x -1)4x + 6y = 22 ( original equation x 2)- 4x + 2y = - 2= 4y = 24÷4y = 62x + 18 = 11-182x = - 7÷2x = - 3.5

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