Gfhghh

6. the arc length formulaL=∫cd(1+(dx/dy)2)dy=∫ab(1+(dy/dx)2)dy

1+(dx/dy)2=1+((y−1)/(2sqrt(y)))2=1+(y−1)2/(4y)=(y2−2y+1+4y)/(4y)=(y+1)2/(4y)

so(1+(dx/dy)2)=((y+1)2/(4y))=(y+1)/2sqrty

L=∫916(y+1)/2sqrt(y)dy=(y+1)/(2sqrty)916=17/8−10/6=0.458

7.f(x)=2x^(3/2)=>f'(x)=3x^(1/2)

L(x)=∫ax(1+(f'(t))2)dt=∫36432(1+9t)dt.Now we can use the substitution rule. Set

u = 1 + 9t and we have∫udu/9=2u(3/2)/27=2(1+9t)(3/2)/27

L(x)=2(1+9t)(3/2)/2736432=17964.83−434=17530.83.

8.y=5cos x where 0<=x<=2pi

L=∫ab(1+(dy/dx)2)dy so dy/dx=-5sin x so L=∫02pi(1+25(sinx)2)dx

9. x=1/3(y^2+2)^3/2 where 2<=y<=3

We use the formula S=∫y1y22piydswhere ds=(1+(dx/dy)2)dy.

ds=sqrt (1+(y(y^2+2)^1/2)^2)=sqrt(1+y^2(y^2+2))=sqrt((y^2+1)^2)=y^2+1

S=∫232piy(y2+1)dy=pi/2(y2+1)223=50pi−12.5pi=37.5pi=117.75.

10.dS = 2pi*y*√[1 + (y')²] dx 

y = 1/4 x² - 1/2 lnx = 1/4*(x² - 2 lnx) 

y' = 1/4*(2x - 2 * 1/x) = 1/2 * (x - 1/x) 

1 + (y')² = 1 + 1/4*(x - 1/x)² = 4/4 + 1/4*(x² - 2 + 1/x²) = 

= 1/4*(4 + x² - 2 +1/x²) = 1/4*(x² + 2 +1/x²) = 1/4*(x + 1/x)² 

So dS = 2pi*y*√[1/4*(x + 1/x)²] dx = 

= 2pi*1/4*(x² - 2 lnx)*1/2*(x + 1/x) dx = 

= pi/4 *(x² - 2 lnx)*(x + 1/x) dx 

S = pi/4*[2, 5] ∫ (x² - 2 lnx)*(x + 1/x) dx =pi/4(-(ln x)^2-2*((x^2 lnx/2)-x^2/4)+x^4/4+x^2/2)[2,5]=(pi/4)*144.13=113.143

3.b) ∫ 3e^(1/x) dx =3( x*e^(1/x) - Ei(1/x) + C )

Thus: 

∫ e^(1/x) dx (from 1 to 2) 
= 3(x*e^(1/x) - Ei(1/x) )(evaluated from 1 to 2) 
= 3[2*e^(1/2) - Ei(1/2)] - [e - Ei(1)] 
≈ 6.060174 

Error=approximation-true value so Error(T4)=|T4- 6.060174|=|6.095679- 6.060174| = approx 0.035505

Error(M4)=|M4-6.060174|=|6.042620-6.060174| = approx 0.017554

c)f(x)=3e^(1/x) , f'(x)=-3e^(1/x)/x^2, f''(x)=(6x+3)e^(1/x)/x^4 which equals 0 for x=-1/2.The maximum value of |f'(x)| where x=-1/2 is  1.624.

ERROR BOUNDS: Suppose |f ′′ (x)| ≤ K for a ≤ x ≤ b. If ET and EM are the errors in the

Trapezoidal and Midpoint Rules, then

|ET | ≤ K(b − a)3 /(12n^2) and |EM | ≤ K(b − a)3 /(24n^2)

In our case a=1,b=2 and K=1.62 so |ET|<=1.62(2-1)^3/(12n^2) so 1.62/(12n^2)<0.0001 so n>36.78 so it's enough to take n=37

|EM|<=1.62/(24n^2) so  1.62/(24n^2)<0.0001 so n>25.98 so it's enough to take n=26

11.Let: 
s be the length of the side of the triangle, 
h be the vertical depth of the trough, 
r be the density of the liquid, 
g be the acceleration due to gravity, 
x be the distance of an infinitesimally thin horizontal slice of liquid from the base of the trough, 
w be the width of the horizontal slice, 
F be the hydrostatic force. 

From the geometry of the equilateral triangle: 
x / w = h / s = sqrt(3) / 2 
h = s sqrt(3) / 2 
w = 2x / sqrt(3) 

dF = rgw(h - x) dx 
F = rg int(0, h) [ w(h - x) ] dx 
= rg int(0, h) [ xs - 2x^2 / sqrt(3) ] dx 
= rg [ sx^2 / 2 - 2x^3 / 3 sqrt(3) ](0, h) 
= rg [ sh^2 / 2 - 2h^3 / 3 sqrt(3) ] 
= rg [ 3s^3 / 8 - s^3 / 4 ] 
= rgs^3(3 / 8 - 1 / 4) 
= rgs^3 / 8 

F = 810 * 9.81 * 6^2 
= 2.86 * 10^5 N.

12.The way to calculate hydrostatic pressure is rho*g*h 

where rho is the density of the liquid = 1000kg/m^3 for water 
g is the acceleration due to gravity, = 9.81 m/sec^2 
h is the depth = 0.7m for (a), and 0.85m for (b), using the average depth of a vertical side 

l=length of a side of the cub

Force is pressure * area, so force is calculated as rho*g*h*area. So the answers are calculated as - 

a) F=PA=pgh*l^2=1000*9.81*0.7*0.3^2 = 618.03N 

b) 1000*9.81*0.85*0.3^2 = 750.465N 

13.I don't have the drawing and the data but suppose its like this one.