Answer: 2r(0)/3.
Step-by-step explanation:
So, we are given one Important data or o or parameter in the question above and that is the function of the form which is given below(that is);
S(r) = (r0 - r) ar^2 -----------------------------(1).
We will now have to differentiate S(r) with respect to r, so, check below for the differentiation:
dS/dr = 2ar (r0 - r ) + ar^2 (-1 ) ---------;(2).
dS/dr = 2ar(r0) - 2ar^2 - ar^2.
dS/dr = - 3ar^2 + 2ar(r0) ------------------(3).
Note that dS/dr = 0.
Hence, - 3ar^2 + 2ar(r0) = 0.
Making ra the subject of the formula we have;
ra[ - 3r + 2r(0) ] = 0. -------------------------(4).
Hence, r = 0 and r = 2r(0) / 3.
If we take the second derivative of S(r) too, we will have;
d^2S/dr = -6ar + 2ar(0). -------------------(5).
+ 2ar(0) > 0 for r = 0; and r = 2r(0)/3 which is the greatest.
Answer:
[tex]r =\frac{2r_{0}}{3}[/tex]
Step-by-step explanation:
We need to take the derivative of S(r) and equal to zero to maximize the function. In this conditions we will find the radius r for which the speed of the air is greatest.
Let's take the derivative:
[tex]\frac{dS}{dr}=a(2r(r_{0}-r)+r^{2}(-1))[/tex]
[tex]\frac{dS}{dr}=a(2r*r_{0}-2r^{2}-r^{2})[/tex]
[tex]\frac{dS}{dr}=a(2r*r_{0}-3r^{2})[/tex]
[tex]\frac{dS}{dr}=ar(2r_{0}-3r)[/tex]
Let's equal it to zero, to maximize S.
[tex]0=ar(2r_{0}-3r)[/tex]
We will have two solutions:
[tex]r = 0[/tex]
[tex]r =\frac{2r_{0}}{3}[/tex]
Therefore the value of r for which the speed of the air is greatest is [tex]r =\frac{2r_{0}}{3}[/tex].
I hope it helps you!
