Volume and leaking liquid
- Thread starter Mickof
- Start date
It is a rectangular box. We have a general formula
[MATH]V = HDW \implies H = \dfrac{V}{DW}.[/MATH]
With me so far?
Now for this specific problem what is the product of D and W? So our formula becomes what?
Now if V[MATH]1[/MATH] and H1 represent the volume and height at the start of an hour and V2 and H2 represent the volume and height the at the end of an hour,
What is H1 in terms of V1?
What is H2 in terms of V2?
What is V1 - V2?
What is H1 - H2?
[MATH]V = HDW \implies H = \dfrac{V}{DW}.[/MATH]
With me so far?
Now for this specific problem what is the product of D and W? So our formula becomes what?
Now if V[MATH]1[/MATH] and H1 represent the volume and height at the start of an hour and V2 and H2 represent the volume and height the at the end of an hour,
What is H1 in terms of V1?
What is H2 in terms of V2?
What is V1 - V2?
What is H1 - H2?
[MATH]\text {If } V = H * D * W, \text { then } V \div D = H * \cancel D * W \div \cancel D \implies V \div D = H * W.[/MATH]
Does that make sense? If two numbers have the same value, then the quotients of dividing by the same divisor must also have the same value.
[MATH]56 = 4 * 14 \implies 56 \div 4 = 14 \implies 14 = 14.[/MATH]
Now do it again
[MATH](V \div D) /div W = (H * \cancel W) \div \cancel W \implies V \div (D * W) = H \implies \dfrac{V}{D * W} = H.[/MATH]
We now have a formula that tells us in general what the relation of height is given volume, depth, and width.
Here width is 100 cm and depth is 60 cm. So if volume is in cubic centimeters, height in centimeters is
[MATH]H = \dfrac{V}{100 * 60} = \dfrac{V}{6000}.[/MATH]
Now with me?
Does that make sense? If two numbers have the same value, then the quotients of dividing by the same divisor must also have the same value.
[MATH]56 = 4 * 14 \implies 56 \div 4 = 14 \implies 14 = 14.[/MATH]
Now do it again
[MATH](V \div D) /div W = (H * \cancel W) \div \cancel W \implies V \div (D * W) = H \implies \dfrac{V}{D * W} = H.[/MATH]
We now have a formula that tells us in general what the relation of height is given volume, depth, and width.
Here width is 100 cm and depth is 60 cm. So if volume is in cubic centimeters, height in centimeters is
[MATH]H = \dfrac{V}{100 * 60} = \dfrac{V}{6000}.[/MATH]
Now with me?
Great.Yes I understand
V=LDH and I can rearrange that to get
H=V/LD and when I sub in values I get
H=V/6000.
just not sure now ?
[MATH]\text {Let } V_1 \text { and } V_2 \text { be respectively the volumes at the start and end of an hour}. [/MATH]
Given the information in the problem, what is [MATH]V_1 - V_2[/MATH] numerically?
What is the conceptual meaning of that difference?
So what is the rate of volume loss per hour? (Be careful with units.)
[MATH]\text {Let } H_1 \text { and } H_2 \text { be respectively the height of oil at the start and end of an hour.}[/MATH]
And given our formulas what do H1 and H2 equal?
Therefore [MATH]H_1 - H_2[/MATH] equals what numerically?
What is the conceptual meaning of that difference?
Well it is not right because you made an arithmetic error. Fix that, and things look reasonable.
You are losing 3600/600000 = 0.6% of the volume per hour
H1 = 600000/6000 = 100.
[MATH]V_2 = 600000- 3600 = 596400 \ne 596,500[/MATH]
[MATH]\therefore H_2 = 596400 \div 6000 = 99.4.[/MATH]
You are losing 100 - 99.4 = 0.6 centimeters per hour, which is 0.6% of of 100. Looks reasonable to me. So what is the loss per minute?
You are losing 3600/600000 = 0.6% of the volume per hour
H1 = 600000/6000 = 100.
[MATH]V_2 = 600000- 3600 = 596400 \ne 596,500[/MATH]
[MATH]\therefore H_2 = 596400 \div 6000 = 99.4.[/MATH]
You are losing 100 - 99.4 = 0.6 centimeters per hour, which is 0.6% of of 100. Looks reasonable to me. So what is the loss per minute?
Yes
[MATH]\dfrac{0.6 \ cm}{\cancel {hr}} * \dfrac{1 \ \cancel {hr }}{60 \ minutes} = \dfrac{0.6 \ cm}{60\ minutes} = 0.01 \ cm \ per \ minute[/MATH]
Do you see why?