Expert: Ahmed Salami - 3/10/2009

Question
QUESTION: find and clasify the critical turnung points of f(x)=cosx+sinx for 0 is less then x and less then 2pi

ANSWER: Hi Phill,
Turning points occur when f'(x) = 0, to classify them we check the value of f''(x).
f''(x) > 0 refers to a minimum point and
f''(x) < 0 refers to a maximum point
For f(x) = cosx + sinx
f'(x) = -sinx + cosx
     = cosx - sinx
equating to zero, we get
cosx = sinx
and in the range 0 < x < 2pi,
x = pi/4 and 5pi/4
but,
f''(x) = -cosx - sinx
f''(pi/4) = -sqrt2   (< 0)
f''(5pi/4) = sqrt2   (> 0)
Therefore, we have a maximum point at x = pi/4 and a minimum point at x = 5pi/4

Regards

---------- FOLLOW-UP ----------

QUESTION: find the area enclosed between the graphs y=x^2 and x=y^2

ANSWER: Hi Phill,
The curves are
y = x^2
and
y = sqrtx
They intercept where x^2 = sqrtx
i.e x = 0 and x = 1
The area between them is therefore the definite integral of y2 - y1 i.e
(sqrtx - x^2) from x = 0 to 1
Therefore,
A = [2x^(3/2)/3  -  x^3/3]  from 0 to 1
 = 1/3

Hope you can figure it all out.

Regards

---------- FOLLOW-UP ----------

QUESTION: the line y=2x is rotated about the the x-axis, between the limits x=0, x=4. Find the the volume of the cone using definite intergration.  

Answer
Hi Phill,
The volume of the cone is the definite integral of #y^2 dx from x = 0 to x = 4, where # represents pi.
Therefore,
V = {#(2x)^2 dx       from x = 0 to x = 4
 = {#4x^2 dx       from x = 0 to x = 4
 = #4x^3/3       from x = 0 to x = 4
 = (#4.4^3/3) - (#4.0^3/3)
 = 256#/3

Regards