Math problem, average rate of change?

Question

I have several problems that I am stuck at, please help me out. Ive tried but cant seem to solve them.


Find the average rate ofchange of the volume of a cube who has side length x (v(x)= x^3) as x changes from 4 to 4.001

Under certain conditions the volume of V of a quantity of air is relted to the pressure p (which is measured in kilopascals) by the equation v(p)=9/p estimate the rate at which the volume is changing at the instant when the pressure is 40 kilopascals. round the answer to four decimals

Answer 1

First one:
The average rate of change of a function f(x) on an interval [a,b] is

[f(b)-f(a)] / (b-a).

In this case, a=4, b=4.001, and f(x)=x^3.

Second one:
Since v(p) = 9/p, the rate at which the volume is changing with respect to pressure is given by dv/dp = -9/p^2, by the power rule, which states that d/dp[p^n] = np^(n-1), with n=-1 in this case. Plug in p=40 and you're done.

Answer 2

Differentiate f'(x) = dv/dx = 3x^2; so that the rate of change at x = 4 is = 3(4^2) = 48 and at x = 4.001 is 3[(4.001)^2] ~ 48 (close to 48); so the average ROC between those two points is about 48.

Same deal, differentiate V'(p) = -9 p^-2; so at V'(p = 40) = - 9/(40^2); you can do the math. Note, the negative sign shows the volume of air decreases under increasing atmospheric pressure.

Answer 3

Differentiating the function g(x) i.e., d/dx (2x - 3) with appreciate to x yields the speed of replace as 2, it is a consistent, and for this reason, there'll be NO replace in g(x), between the above 2 factors. I in difficulty-free words desire, that you've suggested the question wisely; else the above answer, received't be conscious.

Answer 4

Easiest way: Remember that the slope of the line is the rate of change. Find two points on either side of but close to the point of interest and calculate the slope.

Hard way: derivative of the curve at the point, ie, calculus.