Can you please show work? You do not need to answer all the problems even if you know just one your help is appreciated
1) A storage tank in the shape of a right circular cone has a height of 6 meters and a base of radius 2 meters. The cone is inverted and fluid is pumped in at a rate of .002m^3-minute (2 liters/min). At what rate is the level of the fluid rising when the tank is filled to a height of 3meters? Use V= (pie X r squared X h)/3 and R = (H)/3
2) Verify the mean-value theorem for
a) f(x) = x^3 - 3x^2 - 13x + 15 on (-3,1)
(b) f(x) = cos 2x - 2 cos^2 x on ((-pie/2),(pie/4))
3) Locate and classify all critical points, points of inflection, intervals where the Function increases/decreases and where the Function is concave up/down for
a) f(x) = x^4 - 4x^2 + 4x^3 + 1
b) f(x) = (1+ squareroot(x) + cuberoot(x))^9
c) f(x) = x/(ln x)^2
d) f(x) = e^x + e^-2x
e) f(x) = e^(x^3 - x)
4) Using differentials, approximate the numerical value of
b) ln (3), you know ln (e) = 1
c) log (8), you know log (10) = 1
For number 1) I have almost no work. I only drew the Cone out
/ \ the height is 6 and the radius is 2 as you know..but what do I do from there? The only thing I thought of doing was r = h/3 = .002...h = .006...which makes no sense whatsoever.
number 2a) I know I have to use the formula f(b) - f(a)/b-a so
0-0/-4 which is 0/-4....but what do I do with that??? I'm so confused
number 3) I know to find the critical points i have to do f'(x) = 0 and for points of inflection i have to f"(x) = 0 and for increases and decreases and concave up and concave down I just have to plug them in to see + or -...but when I do 3a I get stuck at 4x^3 + 12x^2 - 8x = 0 (the derivative) and then I get that down to 4x(x^2 + 3x - 2)..but how does that factor??
I've tried with great effort to try to do these problems, but the fact remains I just don't understand how to do it. If anyone could help it would be greatly appreciated. I know you probably won't believe me but this is for my own help...I myself don't cheat if someone cheats they are only cheating themselves..whats the point?
I just thought I could give hints for the problem you posted. you are supposed to find the rate at which the level of the fluid is increasing inside the container. Level corresponds to 'h' which means, you gotta find a value of dh/dt. This gives you the rate at which the height of the liquid changes in the container.
V = [pi * r^2* h] /3
You have already got the important substitution to be made. r = h/3
So, V = [pi * (h^2/9) *h ]/3
= [pi * h^3]/ 27
Now, we've expressed the volume entirely as a function of the height. You can follow this up and do the steps given below.
i) By elementary differentiation, find an expression for dv/dt.
ii) This value dv/dt is given in the question as 0.002 m^3/min.
iii) Plug-in other values such as pi and h as given in the question.
iv) Simplify the equation to find the value for dh/dt.
This is quite easy. Hope it helped your preps.
2) a) You were right in your step where you found that [f(b)-f(a)]/b-a = 0. Now, you gotta find a c belonging to [-3,1] such that it f '(c)= 0. f'(x) = 3x^2-6x-13 .
Just solve this equation for the value '0'. i.e 3c^2 - 6c -13 = 0. Basic theory on quadratic equations would help you with the rest of the problem.
b) the 'b' part would be simple as it involves similar steps as to that of 2.a.