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3 years ago

12

After taking the Metro to Dupont Circle in Washington, DC, Jess reached street level by walking up the escalator at a brisk rate

, taking 39 steps during the trip to the top. Suddenly curious about the length of the escalator, Jess returned to the bottom and walked up the same escalator at a leisurely rate, taking steps one fourth as often as on the first trip, taking 12 steps in all. How many steps can be seen on the visible part of the escalator?

1 answer:

pochemuha3 years ago

3 0

Answer:

48

Step-by-step explanation:

If x steps disappear in the time it takes you to briskly walk to the top, then the total number of steps is x+39.

Likewise, at the slower walking rate 4x steps will disappear, so the total number of steps is 4x+12.

These numbers of steps are equal, so we have ...

x +39 = 4x +12

27 = 3x . . . . . . . . . subtract 3x+12

9 = x . . . . . . . . . . . divide by 3

x + 39 = 9 +39 = 48 . . . . The number of visible steps.

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A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito th

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Answer:

Part a)

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b)

Check the attached figure to see the ultimate behavior of the graph.

Part c)

The level = 25, Amplitude = 0.2499

Step-by-step Solution:

Part a)

Given:

$Q(0)=50$

Rate in:

$\frac{1}{4}\left(1+\frac{1}{2}sint\right)\cdot 2\:=\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

Rate out:

$\frac{Q}{100}\cdot 2=\frac{Q}{50}$

So, the differential equation would become:

$\frac{dQ}{dt}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)-\frac{Q}{50}$

Rewriting the equation:

$\frac{dQ}{dt}+\frac{Q}{50}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

As $p(x)$ is the coefficient of $y$ , while $q(x)$ is the constant term in the right side of the equation:

$p\left(x\right)=\frac{1}{50}$

$q\left(x\right)=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

First it is important to determine the function $\mu$ :

$\mu \left(t\right)=e^{\int \:p\left(t\right)dt}$

$=e^{\int \:\left(\frac{1}{50}\right)dt}$

$=e^{\frac{t}{50}}$

The general solution then would become:

$y\left(t\right)=\frac{1}{\mu \left(t\right)}\left(\int \mu \left(t\right)q\left(t\right)dt+c\:\right)$

$=\frac{1}{e^{\frac{t}{50}}}\int e^{\frac{t}{50}}\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)dt+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{1}{e^{\frac{t}{50}}}\left(\frac{-25e^{\frac{t}{50}}\left(50cost-sint\right)}{5002}+25e^{\frac{t}{50}}\right)+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{\left-1250cost+25sint\right}{5002}+25+\frac{1}{e^{\frac{t}{50}}}c$

Evaluate at $t=0$

$50=y\left(0\right)=\frac{\left(-1250cos0+25sin0\right)}{5002}+25+\frac{1}{e^{\frac{0}{50}}}c$

Solve to c:

$c=25+\frac{1250}{5002}$

$\mathrm{Cancel\:}\frac{1250}{5002}:\quad \frac{625}{2501}$

$c=25+\frac{625}{2501}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:25=\frac{25\cdot \:2501}{2501}$

$c=\frac{25\cdot \:2501}{2501}+\frac{625}{2501}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$c=\frac{25\cdot \:2501+625}{2501}$

$c=\frac{63150}{2501}$

$c\approx 25.25$

Therefore, the general solution then would become:

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b) <em>Plot the Solution to see the ultimate behavior of the graph</em>

The graph appears to level off at about the value of $Q=25$ .

The graph is attached below.

Part c)

In the graph we note that the level is $Q=25$ .

Therefore, the level = 25

The amplitude is the (absolute value of the) coefficient of $cost\:t$ in the general solution (as the coefficient of the sine part is a lot smaller):

Therefore,

$A=\frac{1250}{5002}\:\approx 2.499$

Keywords: differential equation, word problem

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<u>Answer:</u>

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Answer:

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