asst 7 quizzes

asst 7 quizzes

Good work on the first question. On the other questions you started well but got a little mixed up on the meaning of the differential. See my notes and send me a revision; it's important for you to know this because it can occur on the test.

here's the lesson 7 quizes. i will send lesson 7 query tomorrow evening and possibly lesson 8. i want to take the test either fri. or mon. i am sending it to both email addresses, along with the query from lesson 6 that i sent already, but hasn't been posted yet.

If a sand pile 3.2 meters high has a mass of 16384 kg, then what would we expect to be the mass of a geometrically similar sand pile 7.2 meters high? If there are 2.9696 billion grains of sand exposed on the surface of the

first sand pile, how many grains of sand we expect to be exposed on the surface of the second?

y(x) = .0031 x^3, vol=y 16384 = k.k (3.2)^3 16384 = k.k (32.768) 16384/32.768=k.k k.k=500.0 so equation is:y(x) = 500 x^3, wght=y y(x) = 500 (7.2)^3 y(x) = 500(373.248) wght=y=186624kilog 2,969,600,000 y'(x) = (186624-16384)/(7.2-3.2)y'(x) = 170240/4 y'(x) = 42560 y'(x) = 3 a x^2. y'(x) = 3 (500) x^2. y'(x) = 1500(2,969,600,000)^2. 13227786240000000000000 this is not right. this can't be what you wanted me to do. i know the surface area rate changes, i know it is at some kind non linear function that increases on a exponential rate, but it's in relation to the angle of the side, so of course it's function for different substances will be a different ratio against the area because of the side angle.

What you have done here is to find the volume x at which the rate of volume change would be `dV / `dx = (186624-16384)/(7.2-3.2). This is sorta close to the idea of the differential estimate, but it doesn't answer the given question.

It's easier to answer the given question. The proportionality for areas is y = k x^2, and one measure of area could be the number of grains visible. So let y = number of grains and x = height, and proceed as in the first question.

If a sand pile 3 meters high has a mass of 50000 kg, then what would we expect to be the mass of a geometrically similar sand pile 17 meters high? Using the differential estimate the mass of sand required to increase the height of the pile from 3 meters to 3.001 meters. y(x) = .0031 x^3, vol=y 50000 = k.k (3.0)^3 50000 = k.k (27) 50000/27=k.k k.k=1851.185185 so equation is:y(x) = 1851.185185 x^3, wght=y y(x) = 1851.85185(17)^3y(x) = 1851.85185(4913) wght=y=9098148kilog

This isn't a differential estimate; this result is in fact the accurate result for the new volume. Comparing with the results of the differential estimate will give you a feel for how close the differential approximation actually is.

The differential is dy = y ' * dx. Using your function y(x) = 1851.85185 x^3, what is y '? What is dx? What therefore is the differential estimate for dy?

How does this estimate compare with the actual change in the volume, which can be found directly from your results?