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Applied Calculus I
Applied Calculus I Major Quiz
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OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va.,
24212-0828 OR email copy of document to dsmith@vhcc.edu,
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- Numerical answers should be correct to 3 significant
figures. You may round off given numerical information to a precision consistent
with this standard.
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questionable parts.
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test pages together.
Test Problems:
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Problem Number 1
Problem: The quadratic depth vs. clock time model corresponding to depths of 54.83482 cm,
38.60514 cm and 29.31098 cm at clock times t = 17.75529, 35.51058 and 53.26587 seconds is depth(t) = .011 t2
+ -1.5 t + 78.
- Show the system of equations we would solve to get this model.
Then use the model to determine the clock time at which depth is 54.3 cm.
Problem: The depth function depth(t) = .011 t2 + -1.5 t + 78 corresponds to
depths of 54.83482 cm, 38.60514 cm and 29.31098 cm at clock times t = 17.75529, 35.51058 and 53.26587 seconds.
- What system of equations would we solve to get this model?
According to the model, at what precise clock time will the rate of depth change be
- .00946 cm / sec?
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Problem Number 2
Problem: Write the differential equation expressing the statement that the rate which
the temperature T changes with respect to time t is proportional to the difference between
the temperature T and the 15 degree room temperature.
Problem: If dy / dt = 1.14 y^2 + 1.22 y/(t+1), and if at t = 0 we have y = .45, then find
the approximate value of y when t = .3. Using the new values of y and t, find
approximate value y when t = .6. Continue for two more steps to find the approximate
value of y when t = 1.2.
(extra credit): Use a predictor-corrector method, with `
Dt = .6 instead of the .3 used above, to find the
approximate value of y when t = 1.2. Which value do you think is more accurate?
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Problem Number 3
The depth vs. clock time function y = .028 t2 + -2.8 t + 90 indicates the depth
y of water in a certain uniform cylinder at clock time t.
- At what average rate does depth changes between clock times t = 13.7 and t = 27.4?
- What clock time lies midway between t = 13.7 and t = 27.4, at what rate is depth changing
at this instant?
What is the function that represents the rate r of depth change at clock time t?
- Evaluate this function at the clock time halfway between t = 13.7 and t = 27.4.
If the rate of depth change is given by dy/dt = .194 t + -2.7 represents the rate at which
depth is changing at clock time t, then how much depth change will there be between clock
times t = 13.7 and t = 27.4?
- Give the function that represents
the depth. What would this specific function be if at clock time t = 0 the depth is 120?
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Problem Number 4
The depth of water in a certain nonuniform container is y = .016 t4 + -1.1 t2
+ 67, where depth y is in cm when clock time t is in seconds.
- At what average rate is the depth of water changing between clock times t = 9.6 and t =
9.700001 seconds?
- At what average rate is the depth of water changing between clock times t = 9.6 and t =
9.610001 seconds?
- At what average rate is the depth of water changing between clock times t = 9.6 and t =
9.601001 seconds?
- What do you estimate is the rate at which water depth is changing at clock time t = 9.6
seconds?
The rate at which water flows from a certain nonuniform cylinder is given by rate = .016
t3 + -1.1 t cm3 per minute, where t is in minutes. How much do
water do you think will flow between clock times t = 9.700001 minutes and t = 19.2 minutes?
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Problem Number 5
Sketch and completely label a
trapezoidal approximation graph for the function y = 4 x .5 + 1, for x = 0 to
2.4 by increments of .8.
