(1) Determine the length of the curve y= 3 + 2x^(3/2) for 0 ≤ x ≤ 2. You can use your calculator at the last step (after the integration) to determine the approximate length, or you may keep it exact.

Please answer all 3 questions and explain . thank you.

(2) Set up, but do not evaluate an integral for the surface area obtained by rotating the curve y = x^3+ 4, 1 ≤ x ≤ 5 (a) about the x-axis. (b)about the y-axis.

(3)Find the particular solution to the differential equation dy dx = y 2 sin x that satisfies the initial condition y(0) = 3.

Expert Solution

milcah answered 1 year ago

Find the exact length of the curve. y = ln (1-x^2), 0 ≤ x ≤ (1/7)

(a) Find the exact length of the curve y = 1/6 (x2 + 4)(3/2) , 0...

(a) Find the exact length of the curve y = 1/6 (x2 + 4)(3/2) , 0 ≤ x ≤ 3. (b) Find the exact area of the surface obtained by rotating the curve in part (a) about the y-axis. I got part a I NEED HELP on part b

y' = 2 + t^2 + y^2 0<t<1 y(0)=0 use the euler method to determine step...

y' = 2 + t^2 + y^2 0<t<1 y(0)=0 use the euler method to determine step size (h) to keep global truncation error below .0001

f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1...

f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1 given X is the number of students who get an A on test 1 given Y is the number of students who get an A on test 2 find the probability that more then 90% students got an A test 2 given that 85 % got an A on test 1

Find the arc length of the curve x^3 = y^2 between x = 1 , x...

Find the arc length of the curve x^3 = y^2 between x = 1 , x = 4 .

Determine the area between the curve f(x)= |2x-3|, the x axis and the lines x=0 and...

Determine the area between the curve f(x)= |2x-3|, the x axis and the lines x=0 and x=3 a. 4.5 b. 2.25 c. 6 d. 3

determine the minimum ououtnkf the function f(x,y)=x^2+y^2-2x-8y on the region specified x>0 0<y<5 y>x

Let Q1=y(2) Q2 =y(3) where y=y(x) solves y'+2xy=2x^3 y(0)=1 dy/dx= (2x-y)/(x+4y) y(1)=1 y'+ycotx=y^3sin^3x y(pi/2)=1 (y^2-2x)dx+2xy dy=0 ...

Let Q1=y(2) Q2 =y(3) where y=y(x) solves y'+2xy=2x^3 y(0)=1 dy/dx= (2x-y)/(x+4y) y(1)=1 y'+ycotx=y^3sin^3x y(pi/2)=1 (y^2-2x)dx+2xy dy=0 y(1)=1 2x^2y;+4xy=3sinx y(2pi)=0 y"+2y^-1 (y')^2=y'

Solve for x x^4 + 2x^3 + 2x^2 + 3x + 1 = 0

2. Given the System of Equations: 3x+2y+z+20w= 6 x+2y+z+10w=0 x+y+z+6w=2 2x+2y+z+15w=3 a) Use your calculator to...

2. Given the System of Equations: 3x+2y+z+20w= 6 x+2y+z+10w=0 x+y+z+6w=2 2x+2y+z+15w=3 a) Use your calculator to solve, leaving solution in parametric form b) Find the specific solution when y = 6 c) Perform, BY HAND, a full check of this particular solution

Question

(1) Determine the length of the curve y= 3 + 2x^(3/2) for 0 ≤ x ≤ 2. You can use your calculator at the last step (after the integration) to determine the approximate length, or you may keep it exact.

Solutions

Expert Solution

Related Solutions

Find the exact length of the curve. y = ln (1-x^2), 0 ≤ x ≤ (1/7)

(a) Find the exact length of the curve y = 1/6 (x2 + 4)(3/2) , 0...

y' = 2 + t^2 + y^2 0<t<1 y(0)=0 use the euler method to determine step...

f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1...

Find the arc length of the curve x^3 = y^2 between x = 1 , x...

Determine the area between the curve f(x)= |2x-3|, the x axis and the lines x=0 and...

determine the minimum ououtnkf the function f(x,y)=x^2+y^2-2x-8y on the region specified x>0 0<y<5 y>x

Let Q1=y(2) Q2 =y(3) where y=y(x) solves y'+2xy=2x^3 y(0)=1 dy/dx= (2x-y)/(x+4y) y(1)=1 y'+ycotx=y^3sin^3x y(pi/2)=1 (y^2-2x)dx+2xy dy=0 ...

Solve for x x^4 + 2x^3 + 2x^2 + 3x + 1 = 0

2. Given the System of Equations: 3x+2y+z+20w= 6 x+2y+z+10w=0 x+y+z+6w=2 2x+2y+z+15w=3 a) Use your calculator to...