can somebody give me a calculus problem and solution please?

Answer 1

A rancher has 300 feet of fencing to enclose a pasture bordered on one side by a river. The river side of the pasture needs no fence. Find the dimensions of the pasture that will produce a pasture with a maximum area.

1. Draw a Picture- make sure you leave one side unbounded(the river side) & dont use it in calculating the perimeter.
2. Label one side x and one side y
3. You only have 300 ft of fencing so 300 is the greatest perimeter.
4. So P: 300=2x+y
5. The area is what you are trying to maximize.
A=xy
Substitute:
A=x(300-2x)
=300x-2x^2
6. Take the derivative of the area. When a derivative is equal to 0 the number is either a max or min.
dA=-4x+300=0
x=75
7. Substitue in dimensions formula
2(75) + y =300
y= 150
8. Dimensions are
75 ft x 150 ft

Answer 2

Here's a simple-looking problem that turns out to be incredibly hard:

∫√tan x dx

Solution:

1/(2√2) ln |tan x - √(2 tan x) + 1| + 1/√2 arctan (√(2 tan x) - 1) - 1/(2√2) ln |tan x + √(2 tan x) + 1| + 1/√2 arctan (√(2 tan x)+1) + C

Method:

Let u=√tan x, du = 1/(2√tan x) * sec² x dx = (u⁴+1)/(2u) dx, dx = 2u/(u⁴+1) du

∫2u²/(u⁴+1) du

Factoring u⁴+1, we see that the roots are e^(±iπ/4) and e^(±3iπ/4), so the real quadratic factors are (u - e^(iπ/4))(u - e^(-iπ/4)) = u² - (e^(iπ/4) + e^(-iπ/4))u + 1 = u² - u√2 + 1, and (u - e^(3iπ/4))(u - e^(-3iπ/4)) = u² - (e^(3iπ/4) + e^(-3iπ/4))u + 1 = u² + u√2 + 1. So we have:

∫2u²/((u² + u√2 + 1)(u² - u√2 + 1)) du

Doing a partial fraction decomposition:

2u²/((u² + u√2 + 1)(u² - u√2 + 1)) = (Au+B)/(u² + u√2 + 1) + (Cu+D)/(u² - u√2 + 1)

2u² = (Au+B)(u² - u√2 + 1) + (Cu+D)(u² + u√2 + 1)

Letting u=0:

0 = B+D

Letting u = e^(iπ/4) = (1+i)/√2:

2i = (C(1+i)/√2+D)(i + (1+i) + 1)
2i√2 = (C + iC + D√2)(2+2i)
2i√2 = 2C + 2iC + 2D√2 + 2iC - 2C + 2iD√2
i√2 = 2iC + D√2 + iD√2

Since this is a decomposition over the reals, C and D are real, so equating real and imaginary parts:

0 = D√2, so D=0, and:

√2 = 2C+D√2 = 2C
C = 1/√2

Finally, we let u=1:

2 = (A+B)(1 - √2 + 1) + (C+D)(1 + √2 + 1)
2 = A(2-√2) + 1/√2 (2+√2)
2 - (2+√2)/√2 = (2-√2)A
2 - (√2+1) = (2-√2)A
1-√2 = (2-√2)A
(1-√2)(2+√2) = 2A
2 - 2√2 + √2 - 2 = 2A
-√2 = 2A
A = -1/√2

So to sum up, we have:

A=-1/√2
B=0
C=1/√2
D=0

Thus we have:

∫2u²/((u² + u√2 + 1)(u² - u√2 + 1)) du
1/√2 ∫u/(u² - u√2 + 1) - u/(u² + u√2 + 1) du

Now, we break up the fractions a little:

1/√2 ∫(u - 1/√2)/(u² - u√2 + 1) + (1/√2)/(u² - u√2 + 1) - (u + 1/√2)/(u² + u√2 + 1) + (1/√2)/(u² + u√2 + 1) du

And splitting this into four integrals:

1/√2 ∫(u - 1/√2)/(u² - u√2 + 1) du + 1/2 ∫1/(u² - u√2 + 1) du - 1/√2 ∫(u + 1/√2)/(u² + u√2 + 1) du + 1/2 ∫1/(u² + u√2 + 1) du

Now, using the substitution v=u²-u√2+1, dv = 2u - √2 du = 2(u - 1/√2) du:

1/(2√2) ∫1/v dv + 1/2 ∫1/(u² - u√2 + 1) du - 1/√2 ∫(u + 1/√2)/(u² + u√2 + 1) du + 1/2 ∫1/(u² + u√2 + 1) du

1/(2√2) ln|v| + 1/2 ∫1/(u² - u√2 + 1) du - 1/√2 ∫(u + 1/√2)/(u² + u√2 + 1) du + 1/2 ∫1/(u² + u√2 + 1) du

1/(2√2) ln |u² - u√2 + 1| + 1/2 ∫1/(u² - u√2 + 1) du - 1/√2 ∫(u + 1/√2)/(u² + u√2 + 1) du + 1/2 ∫1/(u² + u√2 + 1) du

Now using the substitution v = u²+u√2+1, dv = 2(u+1/√2) du on the third integral:

1/(2√2) ln |u² - u√2 + 1| + 1/2 ∫1/(u² - u√2 + 1) du - 1/(2√2) ∫1/v dv + 1/2 ∫1/(u² + u√2 + 1) du

1/(2√2) ln |u² - u√2 + 1| + 1/2 ∫1/(u² - u√2 + 1) du - 1/(2√2) ln |v| + 1/2 ∫1/(u² + u√2 + 1) du

1/(2√2) ln |u² - u√2 + 1| + 1/2 ∫1/(u² - u√2 + 1) du - 1/(2√2) ln |u² + u√2 + 1| + 1/2 ∫1/(u² + u√2 + 1) du

Now, completing the square on both of the remaining integrals:

1/(2√2) ln |u² - u√2 + 1| + 1/2 ∫1/(u² - u√2 + 1/2 + 1/2) du - 1/(2√2) ln |u² + u√2 + 1| + 1/2 ∫1/(u² + u√2 + 1/2 + 1/2) du

1/(2√2) ln |u² - u√2 + 1| + 1/2 ∫1/((u - 1/√2)² + 1/2) du - 1/(2√2) ln |u² + u√2 + 1| + 1/2 ∫1/((u + 1/√2)² + 1/2) du

1/(2√2) ln |u² - u√2 + 1| + ∫1/((u√2 - 1)² + 1) du - 1/(2√2) ln |u² + u√2 + 1| + ∫1/((u√2 + 1)² + 1) du

Now, making in the second integral the substitution v = u√2-1, dv = √2 du, and in the fourth integral w = u√2+1, dw = √2 du, we have:

1/(2√2) ln |u² - u√2 + 1| + 1/√2 ∫1/(v² + 1) dv - 1/(2√2) ln |u² + u√2 + 1| + 1/√2 ∫1/(w² + 1) dw

1/(2√2) ln |u² - u√2 + 1| + 1/√2 arctan v - 1/(2√2) ln |u² + u√2 + 1| + 1/√2 arctan w + C

1/(2√2) ln |u² - u√2 + 1| + 1/√2 arctan (u√2-1) - 1/(2√2) ln |u² + u√2 + 1| + 1/√2 arctan (u√2+1) + C

Now, finally using the fact that u=√tan x:

1/(2√2) ln |tan x - √(2 tan x) + 1| + 1/√2 arctan (√(2 tan x) - 1) - 1/(2√2) ln |tan x + √(2 tan x) + 1| + 1/√2 arctan (√(2 tan x)+1) + C

And we are done.

Answer 3

calculus is a big ocean with many sub topics
you better specify the area in which you want the problem

Answer 4

sure!

Find the derivative:
f(x) = ^6 (sq rt (x))

Solution:
f(x) = ^6 (sq rt (x)) = x^(1/6)
f'(x) = (1/6)x^(-5/6) = 1 / (6x^(5/6))