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ehidna [41]
2 years ago
8

A figure is located at (0, 0), (−3, −4), and (−3, 0) on a coordinate plane. What kind of 3-D shape would be created if the figur

e was rotated around the x-axis? Provide an explanation and proof of your answer to receive full credit. Include the dimensions of the 3-D shape in your explanation.
Mathematics
1 answer:
AysviL [449]2 years ago
8 0

Answer:

cone

Step-by-step explanation:

Define the given points:

  • A = (0, 0)
  • B = (-3, -4)
  • C = (-3, 0)

As two points have the same x-coordinate and two different points have the same y-coordinate, the <u>2D shape</u> the points make is a right triangle (see first attached diagram).

As points B and C lie on the <u>same vertical line</u> (x = -3), and point C lies on the x-axis, <u>rotating</u> line segment BC about the x-axis will produce a circle with center C and radius BC (4 units).

As point A lies on the x-axis, it will <u>not move</u> when rotated about the x-axis.

Therefore, the 3D shape the triangle will create is a cone with:

  • vertex at (0, 0)
  • center of circular base at (-3, 0)
  • radius of circular base of 4 units
  • height of 3 units

(see second attached diagram)

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Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

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Answer:

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