asst 6

asst 6

See my notes, which might well raise as many questions as they answer. Let me know if you need me to clarify any of the specifics.

۫[g assignment #006 vH{߁P Calculus I 04-29-2005

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18:39:40 Query class notes #06 If x is the height of a sandpile and y the volume, what proportionality governs geometrically similar sandpiles? Why should this be the proportionality?

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RESPONSE --> both, because they are both used in a ratio or graph to define the geometric shape, along with the slope of the sides, pretty sure this ties into the golden ratio somehow...

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18:40:32 ** the proportionality is y = k x^3. Any proportionality of volumes is a y = k x^3 proportionality because volumes can be filled with tiny cubes; surface areas are y = k x^2 because surfaces can be covered with tiny squares. **

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RESPONSE --> so thats what you wanted...

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18:47:28 If x is the radius of a spherical balloon and y the surface area, what proportionality governs the relationship between y and x? Why should this be the proportionality?

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RESPONSE --> i know it's got pi & an exponent in it, but i cant remember the formula.....pi r^3 ?

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18:50:30 ** Just as little cubes can be thought of as filling the volume to any desired level of accuracy, little squares can be thought of as covering any smooth surface. Cubes 'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2.

Surfaces can be covered as nearly as we like with tiny squares (the more closely we want to cover a sphere the tinier the squares would have to be). The area of a square is proportional to the square of its linear dimensions. Radius is a linear dimension. Thus the proportionality for areas is y = k x^2.

By contrast, for volumes or things that depend on volume, like mass or weight, we would use tiny cubes to fill the volume. Volume of a cube is proportional to the cube of linear dimensions. Thus the proportionality for a volume would be y = k x^3. **

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RESPONSE --> ok. but a balloon is a sphere, so wheres pi?

Good question. The answer is that pi is in the proportionality constant k.

The formula for the area of a sphere is 4 pi r^2. So if y is area and x is radius, you have y = k x with k = 4 pi.

If x is the circumference, you would still have y = k x^2 but now k would be different. That is, area has a different relationship with circumference than with radius, even though the form of the proportionality is the same. Specifically, radius = circumference / (2 pi) so the area would be 4 pi r^2 = 4 pi ( circumference / (2 pi) ) ^ 2 = circumference^2 / pi. So if y is area and x is circumference, you have y = k x^2 but with k = 1 / pi.

The point here is that you don't need to even know the formula for the area, provided you know the area and, say, the diameter of the circle. For example if the diameter is 10 and the area is 314, you let y be area and x be diameter. Your proportionality is y = k x^2, with y = 314 when x = 10. So k = y / x^2 = 314 / 10^2 = 3.14, so your proportionality becomes y = 3.14 x^2 and now you can find the area for any diameter.

If you know the area and circumference you can follow the same procedure, though you will get a different value of k. For example if area is 314 when the circumference is 31.4, then you write y = k x^2 for x = 31.4 and y = 314 and you get k = y / x^2 = 314 / (31.4)^2 = .31, approx., and your proportionality is y = .31 x^2. Plug in any circumference x and you get the area of the sphere.

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19:15:37 Explain how you would use the concept of the differential to find the volume of a sandpile of height 5.01 given the volume of a geometrically similar sandpile of height 5, and given the value of k in the y = k x^3 proportionality between height and volume.

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RESPONSE --> y=kx^3 y=1.01(5.01 )^3 y=this is not right/i give

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20:29:14 ** The class notes showed you that the slope of the y = k x^3 graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the given information, you can evaluate y' at x = 5. That gives you the slope of the line tangent to the curve, and also the rate at which y is changing with respect to x. When you multiply this rate by the change in x, you get the change in y.

The differential is 3 k x^2 `dx and is approximately equal to the corresponding `dy. Since `dy / `dx = 3 k x^2, the differential looks like a simple algebraic rearrangement `dy = 3 k x^2 `dx, though what's involved isn't really simple algebra. The differential expresses the fact that near a point, provided the function has a continuous derivative, the approximate change in y can be found by multiplying the change in x by the derivative). That is, `dy = derivative * `dx (approx)., or `dy = slope at given point * `dx (approx), or `dy = 3 k x^2 `dx (approx).

The idea is that the derivative is the rate of change of the function. We can use the rate of change and the change in x to find the change in y.

The differential uses the fact that near x = 5 the change in y can be approximated using the rate of change at x = 5.

Our proportionality is y = k x^3. Let y = f(x) = k x^3. Then y' = f'(x) = 3 k x^2. When x = 5 we have y' = f'(5) = 75 k, whatever k is. To estimate the change in y corresponding to the change .01 in x, we will multiply y ' by .01, getting a change of y ' `dx = 75 k * .01. } SPECIFIC EXAMPLE: We don't know what k is for this specific question. As a specific example suppose our information let us to the value k = .002, so that our proportionality is y = .002 x^3. Then the rate of change when x is 5 would be f'(5) = 3 k x^2 = 3 k * 5^2 = 75 k = .15 and the value of y would be y = f(5) = .002 * 5^3 = .25. This tells us that at x = 5 the function is changing at a rate of .15 units of y for each unit of x.

Thus if x changes from 5 to 5.01 we expect that the change will be

change in y = (dy/dx) * `dx = rate of change * change in x (approx) = .15 * .01 = .0015,

so that when x = 5.01, y should be .0015 greater than it was when x was 5. Thus y = .25 + .0015 = .2515. This is the differential approximation. It doesn't take account of the fact that the rate changes slightly between x=5 and x = 5.01. But we don't expect it to change much over that short increment, so we expect that the approximation is pretty good.

Now, if you evaluate f at x = 5.01 you get .251503. This is a little different than the .2515 approximation we got from the differential--the differential is off by .000003. That's not much, and we expected it wouldn't be much because the derivative doesn't change much over that short interval. But it does change a little, and that's the reason for the discrepancy. The differential works very well for decently behaved functions (ones with smooth curves for graphs) over sufficiently short intervals.**

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RESPONSE --> oookkk...the left hemisphere is now melted into the right. one would think that theoretically the brain would work better, but it don't. i THINK i got it.

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21:32:54 What would be the rate of depth change for the depth function y = .02 t^2 - 3 t + 6 at t = 30? (instant response not required)

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RESPONSE --> y = .02 (30)^2 - 3 (30) + 6 y = .02 (900) - 90 + 6 y = 18 - 96 y = -78 ok. i give up.i cannot understand t2 = y(t2) - y(t1) = (.5 m t2^2 + b t2 + c) - (.5 m t1^2 + b t1 + c or `dy / `dt = [ y(t + `dt) - y(t) ] / `dt, for y(t) i plugged it in, and got: 'y(30t-46800 )

See my note below, but check out the algebra here, since you were on the right track.

First note that since you are given y in the form a t^2 + b t + c, you would be better off using the form y ' = 2 a t + b.

However, the form .5 m t^2 + b t + c works too, if as you do below you assume derivative y ' = m t + b.

If you are using t1 and t2, as opposed to t and t + `dt, then `dt = t2 - t1.

In this case

`dy / `dt = [ (.5 m t2^2 + b t2 + c) - (.5 m t1^2 + b t1 + c) ] / (t2 - t1).

It takes a little algebra but this reduces to [ .5 m ( t2 + t1)( t2 - t1) + b ( t2 - t1) ] / (t2 - t1); the t2 - t1 divides both terms of the numerator leaving you

.5 m ( t2 + t1) + b.

As `dt -> 0, t2 - t1 -> 0 so t2 -> t1. This means that t2 + t1 -> 2 t1 and we have

.5 m ( 2 t1) + b = m t1 + b.

taking the limit as `dt approaches zero- how do i do that-? also, i thought i didn't need the (or wasn't supposed to) do the trapazoid to determine distance? if you could show it in(baby-?) steps for the derivitive, i'd probably get it.

You should know how, but you don't have to do the limit definition every time.

It was shown in an earlier assignment that if you do the limit definition for y = a t^2 + b t + c, you get y ' = 2 a t + b. You can use that formula as follows:

For this one a = .02, b = 3 so y ' = 2 a t + b = 2 * .02 t + 3 = .04 t + 3.

Plugging in t = 30 we get

y ' = .04 * 30 + 3 = 4.2. That is the rate at the t = 30 instant.

Alternatively, using the fact that y = .5 m t^2 + b t + c corresponds y ' = m t + b, you see that .5 m in front of the t^2 tells you that .5 m = .02 so m = .04. You come to the same conclusion: y ' = .04 t + 3.

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