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Mth 174

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Work to Lift Weight With Chain

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Problem Number 6

Use Riemann Sums to obtain the integral required to solve the following, and evaluate the integral: How much work is required to lift a weight of 300 pounds a vertical distance of 23 feet using a chain that weighs 3 lb/ft?

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F(x) = sum( weight of chain + weight of the weight)

= 3(23 - x) + 300

= 69 - 3x + 300

= 369 - 3x

W = integral(369 - 3x dx, x, 0, 23)

= 369x - 3 x^2 from 0 to 23

= 6900 ft/lb

Is this solution correct?

@&

You have the right integral but no Riemann Sum.

The typical interval has sample point x_i* and length `dx.

When the weight is at position x_i*, the length of chain below the 23 foot mark is 23 ft - x_i*, so the weight of that section is 3 lb / ft * (23 ft - x_i*) and the work to raise it through the increment `dx is

`dW_i = (300 + 3 ( 23 - x_i*) ) * `dx

where the units have not been included.

The Riemann sum is therefore

sum( (300 + 3(23 - x_i*)) `dx)

which approaches the integral you have given, except that you need an additional sign of grouping:

W = integral( (369 - 3x) dx, x, 0, 23)

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question form

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Mth 174

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Temperature Change Question

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Problem Number 7

If an object's temperature moves 35% closer to the 24 degree temperature of a room in any 6-minute time period, then if the object's initial temperature is 42.8 degrees, what function models the temperature as a function of time?  What will be the temperature after 34 minutes?

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I'm not sure what the differential equation is for this problem?

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The rate at which and object's temperature approaches room temperature is proportional to the difference between the object's temperature and that of the room.

That rate is dT/dt.

The temperature difference is (T - T_room).

So the equation is

dT/dt = k (T - T_room).

If you solve the equation, which you can easily do using separation of variables, you get

T = T_room + C e^(kt)

The given conditions allow you to solve for C and k.

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