(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 2 years 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

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

Determine if exact equation, then solve: x(y+2)y' = lnx+1 , y(1) = -1

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)

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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

Determine if exact equation, then solve: x(y+2)y' = lnx+1 , y(1) = -1