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| − | <span class="exam">Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure <math style="vertical-align: 0px">P</math> and volume <math style="vertical-align: 0px">V</math> satisfy the equation <math style="vertical-align: 0px">PV=C</math> where <math style="vertical-align: 0px">C</math> is a constant. Suppose that at a certain instant, the volume is <math style="vertical-align: -4px">600 \text{ cm}^3,</math> the pressure is <math style="vertical-align: -4px">150 \text{ kPa},</math> and the pressure is increasing at a rate of <math style="vertical-align: -4px">20 \text{ kPa/min}.</math> At what rate is the volume decreasing at this instant? | + | <span class="exam">If <math style="vertical-align: 0px">W</math> denotes the weight in pounds of an individual, and <math style="vertical-align: 0px">t</math> denotes the time in months, then <math style="vertical-align: -13px">\frac{dW}{dt}</math> is the rate of weight gain or loss in lbs/mo. The current speed record for weight loss is a drop in weight from 487 pounds to 130 pounds over an eight month period. Show that the rate of weight loss exceeded 44 lbs/mo at some time during the eight month period. |
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| | + | [[009A Sample Final 3, Problem 8 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |'''Product Rule'''
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| − | | <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
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| − | |}
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| | + | [[009A Sample Final 3, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |First, we take the derivative of the equation <math style="vertical-align: 0px">PV=C.</math>
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| − | |Using the product rule, we get
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| − | | <math>P'V+PV'=C'.</math>
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| − | |Since <math style="vertical-align: 0px">C</math> is a constant,
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| − | | <math style="vertical-align: -1px">C'=0.</math>
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| − | |Therefore, we have
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| − | | <math>P'V+PV'=0.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Solving for <math style="vertical-align: -4px">V',</math> we get
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| − | | <math>V'=\frac{-P'V}{P}.</math>
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| − | |Using the information provided in the problem, we have
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| − | | <math style="vertical-align: 0px">V=600 \text{ cm}^3,~P=150 \text{ kPa},~P'=20 \text{ kPa/min}.</math>
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| − | |Hence, we get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{V'} & = & \displaystyle{\frac{-(20)(600)}{150} \text{ cm}^3\text{/min}}\\
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| − | &&\\
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| − | & = & \displaystyle{-80 \text{ cm}^3\text{/min}.}
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| − | \end{array}</math>
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| − | |Therefore, the volume is decreasing at a rate of <math style="vertical-align: -5px">80 \text{ cm}^3\text{/min}</math> at this instant.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | The volume is decreasing at a rate of <math style="vertical-align: -5px">80 \text{ cm}^3\text{/min}</math> at this instant.
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| − | |}
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| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
