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avanturin [10]
2 years ago
9

A cylinder has a diameter of 6 feet. the volume of the cylinder is 36/5π cubic feet.

Mathematics
1 answer:
Anton [14]2 years ago
7 0

Answer:

Height of cylinder is \frac{4}{5}\ ft.

Step-by-step explanation:

Diameter of cylinder, D = 6 feet

Relation between Radius, R and diameter, D is given by:

R = \dfrac{D}{2}

So, radius, R = \frac{6}{2} \Rightarrow 3 feet

Let height of cylinder be h feet.

Formula for volume of cylinder is given by:

V = \pi R^{2} H

Where, R is the radius of base of cylinder

and H is the height of cylinder

We are given that,

V = \dfrac{36 \pi}{5} ft^{3}

Using formula for volume, V = \pi R^{2} H

\dfrac{36\pi}{5} = \pi \times 3^{2} \times h\\\Rightarrow \dfrac{36\pi}{5} = \pi \times 9 \times h\\\Rightarrow h = \dfrac{4}{5} ft

Hence, Height of cylinder is \frac{4}{5}\ ft.

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Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] < y0 < [infinity](a) Sketch the graph
kotegsom [21]

Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of  f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0  and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane  is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of  f(y) versus y

Looking at the given differential equation

       \frac{dy}{dt} = e^{y} - 1 for -∞ < y_{o} < ∞

 We can say let \frac{dy}{dt} = f(y) =e^{y} - 1

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

Step Two : Determine the critical point

   To fin the critical point we have to set   \frac{dy}{dt} = 0

       This means e^{y} - 1 = 0

                          For this to be possible e^{y} = 1

                          which means that  e^{y} = e^{0}

                          which implies that y = 0

Hence the critical point occurs at y = 0

meaning that the equilibrium solution is y = 0

As t → ∞, our curve is going to move away from y = 0  hence it is asymptotically unstable.

Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where  f(y) = 0.

In each of the intervals bounded  by the equilibrium draw an upward

pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.

      This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For  below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper  

Step Four : Draw a Solution Curve

A solution curve is a curve that shows the solution of a DE (deferential equation)

Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium  that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)

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Answer:

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Step-by-step explanation:

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