Question

Snow is piling on a driveway so its depth is changing at a rate of r(t) = 10sqrt(1 - cos(0.5t)) centimeters per hour, where t is the time in hours, 0 <= t <= 5 At time t = 0 , the depth of the snow is 20 centimeters. Use a graphing calculator and round your answer to three decimal places. What is the snow's depth at time t = 5 hours? centimeters

Answer 1

58.731 cm

Explanation:

You want the depth of accumulated snow after 5 hours if it is 20 cm at 0 hours and is increasing depth at the rate r(t) = 10√(1 -cos(0.5t)) cm/hour.

Integration

The attached calculator screen shows the result of the integration. The mode is set to RADians. It took this calculator just over 1 minute to provide its result to 13 significant figures.

The snow's depth at t = 5 hours is 58.731 cm.

__

Additional comment

A different graphing calculator provided the same result to 12 significant figures "instantly." That is shown in the second attachment.

When scientific calculators were rare and expensive, their capability for numerical integration was more limited. Generally one had to specify the tolerance the calculator was to use for its result. These calculators both provided the result to full calculator precision.

The exact result is 20+80(√2)sin(5/8)² ≈ 58.731216038167283412... cm.

We're not quite sure the point of an answer precise to 0.001 cm. The diameter of a snowflake may range from 0.050 to 0.500 cm.