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# Tutor profile: William B.

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William B.
Software Engineer and all around Math Geek
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## Questions

### Subject:Calculus

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Question:

Find the curve length of $$y = x^\frac{3}{2}$$ from x = 0 to x = 4.

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William B.

Arc length is given by the formula: $$S = \int_{a}^{b}\sqrt{1+f'(x)^2}dx$$ Since $$f'(x) = \frac{3}{2}x^\frac{1}{2}$$ We can set up $$S = \int_{0}^{4}\sqrt{1+[\frac{3}{2}x^\frac{1}{2}]^2}dx = \int_{0}^{4}\sqrt{1+\frac{9}{4}x}dx$$ We can use u substitution like this: $$u = 1 + \frac{9}{4}x$$ $$du = \frac{9}{4}dx$$ $$\frac{4}{9}du = dx$$ With new bounds a -> 1 + 0 = 1 b -> 1 + 9/4*4 = 10 $$\int_{1}^{10}\sqrt{u}*\frac{4}{9}du = \frac{4}{9}\int_{1}^{10}u^\frac{1}{2}du$$ Finally: $$S = \frac{4}{9} [\frac{2}{3}(10)^\frac{3}{2}-\frac{2}{3}(1)^\frac{3}{2}] = 9.073$$

### Subject:Physics

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Question:

Legolas pulls his bow back 0.8m which has a stiffness of 180 N/m. His arrow weighs 45 g. Once the arrow gets up to speed, what is the velocity of the arrow?

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William B.

We could integrate the force over the bow but what is even easier is to just use the energy equation. Any linear force field (spring, bow, etc) holds potential energy in the equation $$PE = \frac{1}{2}kx^2$$ Where x is the distance traveled and k is the "spring" constant. This comes out to $$PE = \frac{1}{2}(180)(.8)^2 = 57.6J$$ Now we will assume that all of this energy gets converted into kinetic energy. This is given by $$KE = \frac{1}{2}mv^2$$ where m is the mass of the object and v is its velocity Therefore we isolate for v $$v = \sqrt\frac{2KE}{m} = \sqrt\frac{2*57.6J}{.045kg} = 50.6 m/s$$

### Subject:Algebra

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Question:

At the Wonka Chocolate factory there is a 100 gallon vat of liquid chocolate that contains 90% chocolate (and 10% milk). One of the oompa-loompas didn't read the recipe and turned on the nozzle that poured in a 50% chocolate mixture (with 50% milk) at a rate of 5 gallons/min. At the same time, the whole vat is being emptied into the candy bars at a rate of 5 gallons per minute. Assuming the tank is being constantly stirred, what is the percentage of chocolate in the tank after 10 minutes?

Inactive
William B.

What's key here is to track the amount of chocolate in the tank over time. This is given by: Amount = Volume * Concentration. Since the volume of the vat is staying constant, we can just set a relationship to the amount of chocolate in the tank with time. A = 100(.90) + 5(.5)*t - 5(A)/100*t When time = 0 the amount of chocolate is 90 gallons. For each minute that goes by, 5 gallons of 50% mixture is added to that amount. The insight here is to realize the outflow of chocolate is related to the amount inside the tank. We are subtracting 5 gallons every minute, of whatever the amount is inside the tank divided by the total volume. This gives us out whole equation that we can then solve for A. A = (90 + 2.5*t)/(1 + t/20) After 10 minutes we get A = 76.66 gallons, or ~76% chocolate in the tank.

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