Calculate the first AND second derivative dy/dx and d^2y/dx^2 for the curve given by:
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Calculate the first AND second derivative dy/dx and d^2y/dx^2
for the curve given by:
r(t) =...
1a) Find dy/dx x = te', y = t + sin t b) Find dy/dx and...
1a) Find dy/dx x = te', y = t + sin t b) Find dy/dx and d’y/dx2 for which t is curve concave upward x = x3 + 1, y = t - c)Find the points on the curve where the tangent is horizontal or vertical. Draw the graph x = 13 – 3t, y = t3 – 312 d) Find the area enclosed by the x-axis and the curve x = t3 + 1, y = 2t – t?....
Given the System described by the differential equation below: D^2 y(t) + 3 Dy(t) + 2y(t)...
Given the System described by the differential equation below: D^2 y(t) + 3 Dy(t) + 2y(t) = x(t) where D=d/dt and D^2 is the second derivative 1) find its Transfer Function H(s) (assume all initial conditions are zero) 2) use the first procedural way of getting A, B, C, D from H(s) to find the corresponding state space representations . Then do the reverse step of finding H(s) from the A, B, C, D representation just found (i.e. check that...
For the following differential equation: (x^3)dy/dx+y^4+3=0 where dy/dx is the first derivative of y with respect...
For the following differential equation: (x^3)dy/dx+y^4+3=0 where dy/dx is the first derivative of y with respect to x, () means power. The equation has initial values y=2.00 at x=1.00 Using Euler method with a step in the x direction of h=0.30: Show the equation to use to generate values of (2 marks) Calculate the missing values of y in the table below I .1.30 1.00 2.00 1.60 For (2 marks)
Use the given transformation to evaluate the integral. -5x dx dy where R is the parallelogram bou...
Use the given transformation to evaluate the integral. -5x dx dy where R is the parallelogram bounded by the linesy-x+1, y-x +4 y#2x+2y»2x + 5 A) -5 B) 10 C)5 D)-10 32) y+ x where R is the trapezoid with vertices at (6,0), ,0).。. 6), (0.9) 45 45 B) ÷ sin l 45 C) sin 2 45 A) sin 2
Use the given transformation to evaluate the integral. -5x dx dy where R is the parallelogram bounded by the linesy-x+1,...
Dx+Dy = e b. -D r + Dx + I + y = 0 C Dx+Dy = e b. -D r + Dx + I + y = 0 C
Dx+Dy = e b. -D r + Dx + I + y = 0 C
Dx+Dy = e b. -D r + Dx + I + y = 0 C
Solve the initial value problem dy dx+2y-4e0y(O)2 The solution is y(x) Solve the initial value problem dy dx+2...
Solve the initial value problem dy dx+2y-4e0y(O)2 The solution is y(x)
Solve the initial value problem dy dx+2y-4e0y(O)2 The solution is y(x)
Find the particular antiderivative of the following derivative that satisfies the given condition. dy dx -3...
Find the particular antiderivative of the following derivative that satisfies the given condition. dy dx -3 = 2x + 8x - 1; y(1) = 0 y(x) =
Solve for y(t). dy/dt + 2x = et dx/dt-2y= 1 +t when x(0) = 1, y(0)...
Solve for y(t). dy/dt + 2x = et dx/dt-2y= 1 +t when x(0) = 1, y(0) = 2
(d) The line integral [(x+y?)dx + (x2 + 2xy)dy, where the positively oriented curve C is...
(d) The line integral [(x+y?)dx + (x2 + 2xy)dy, where the positively oriented curve C is the boundary of the region in the first quadrant determined by the graphs of x=0, y=x2 and y=1, can be converted to A 2xdydx 0 0 BJ 2 xdxdy 0 0 С -2x)dyda 00 D none of the above (e) Consider finding the maximum and minimum values of the function f(x, y) = x + y2 - 4x + 4y subject to the constraint...
(1 point) Given R(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tkR(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tk Find the derivative R′(t)R′(t) and norm of the derivative. R′(t)=R′(t)= ∥R′(t)∥=‖R′(t)‖= Then...
(1 point)
Given
R(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tkR(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tk
Find the derivative R′(t)R′(t) and norm of the derivative.
R′(t)=R′(t)= ∥R′(t)∥=‖R′(t)‖=
Then find the unit tangent vector T(t)T(t) and the principal
unit normal vector N(t)N(t)
T(t)=T(t)= N(t)=N(t)=
(1 point) Given R(t) = cos(36) i + e sin(3t) 3 + 3e"k Find the derivative R') and norm of the derivative. R'(t) = R' (t) Then find the unit tangent vector T(t) and the principal unit normal vector N() T(0) N() Note: Yn can can on the hom
1a) Find dy/dx x = te', y = t + sin t b) Find dy/dx and d’y/dx2 for which t is curve concave upward x = x3 + 1, y = t - c)Find the points on the curve where the tangent is horizontal or vertical. Draw the graph x = 13 – 3t, y = t3 – 312 d) Find the area enclosed by the x-axis and the curve x = t3 + 1, y = 2t – t?....
For the following differential equation: (x^3)dy/dx+y^4+3=0 where dy/dx is the first derivative of y with respect to x, () means power. The equation has initial values y=2.00 at x=1.00 Using Euler method with a step in the x direction of h=0.30: Show the equation to use to generate values of (2 marks) Calculate the missing values of y in the table below I .1.30 1.00 2.00 1.60 For (2 marks)
Use the given transformation to evaluate the integral. -5x dx dy where R is the parallelogram bounded by the linesy-x+1, y-x +4 y#2x+2y»2x + 5 A) -5 B) 10 C)5 D)-10 32) y+ x where R is the trapezoid with vertices at (6,0), ,0).。. 6), (0.9) 45 45 B) ÷ sin l 45 C) sin 2 45 A) sin 2
Use the given transformation to evaluate the integral. -5x dx dy where R is the parallelogram bounded by the linesy-x+1,...
Dx+Dy = e b. -D r + Dx + I + y = 0 C
Dx+Dy = e b. -D r + Dx + I + y = 0 C
Solve the initial value problem dy dx+2y-4e0y(O)2 The solution is y(x)
Solve the initial value problem dy dx+2y-4e0y(O)2 The solution is y(x)
Find the particular antiderivative of the following derivative that satisfies the given condition. dy dx -3 = 2x + 8x - 1; y(1) = 0 y(x) =
(d) The line integral [(x+y?)dx + (x2 + 2xy)dy, where the positively oriented curve C is the boundary of the region in the first quadrant determined by the graphs of x=0, y=x2 and y=1, can be converted to A 2xdydx 0 0 BJ 2 xdxdy 0 0 С -2x)dyda 00 D none of the above (e) Consider finding the maximum and minimum values of the function f(x, y) = x + y2 - 4x + 4y subject to the constraint...
(1 point)
Given
R(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tkR(t)=e4tcos(3t)i+e4tsin(3t)j+3e4tk
Find the derivative R′(t)R′(t) and norm of the derivative.
R′(t)=R′(t)= ∥R′(t)∥=‖R′(t)‖=
Then find the unit tangent vector T(t)T(t) and the principal
unit normal vector N(t)N(t)
T(t)=T(t)= N(t)=N(t)=
(1 point) Given R(t) = cos(36) i + e sin(3t) 3 + 3e"k Find the derivative R') and norm of the derivative. R'(t) = R' (t) Then find the unit tangent vector T(t) and the principal unit normal vector N() T(0) N() Note: Yn can can on the hom
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