Test
Unit 2 test
Este test tiene tiempo límite, así que se abre en el reproductor de siempre.
Preguntas del test (16)
1find the partial derivitive of this function. f(x,y) = x^2 + y^2
- Adf/dx = 2x + y^2, df/dy = x^2 + 2y
- Bdf/dx = x^2 + 2y, df/dy = 2x + y^2
2What is the gradient of the function? x^3 + y^2
- A
- B
- A
3the directional derivitive of this gradient is [4,3] when the vector = (1,1)?
- AVerdadero
- BFalso
4which paremtric function will draw fastest?
- A2x + 2y
- B3x + 3y
- C(1/2)x + (1/2)x
- D4x + 4y
5what is the location of the point where T = 4 in this function. (1/2)t^2 + (1/2)t^2
- A(8,8)
- B(16,16)
- C(4,4)
- D(0,0)
6What is the derivitive of this functions? f(x,y) = x^2 + y^2 and x(t) = cos(t) and y(t) = sin(t)
- A-2sin(t)(cos(t)(cos^2(t))) + cos^2(t)(sin(y)(2cos(t)))
- B-sin^2(t)(cost)(2cos(t))) + 2cos(t)(sin(t)(cos^2(t)))
7curvture is where you are driving along the road and your wheel locks up, and the curture is the radius of the circle.
- AVerdadero
- BFalso
8find the curvture. f(t) = cos(t) and f(t) = sin(t) and f(t) = t/10
- A99/100
- B101/100
9find the derivitive. f(t,s) = t^2 + s^2 f(t,s) = st f(t,s) ts^2 + st^2
- Adv/dt = 2t, s, s^2 - 2st. dv/ds dv/ds = -2s, t, 2ts - t^2
- Bdv/dt = -2s, t, 2ts - t^2. dv/ds = 2t, s, s^2 - 2st
10when the density increases it is what.
- Anegative divergence
- Bpositive divergence
- Cno divegence
11find the divergence of this function. f(x,y) = x^2 + y^2 and f(x,y) = x^3 + y^3
- A2x + y^2 + x^3 + 3y
- Bx^2 + 2y + 3x + y^3
12what is the curl of this function. f(x,y) = x^2 + y^2 and f(x,y) = x^3 - y^3
- A2x + y^2 - (x^3 - 3y)
- B2x + Y^2 + x^3 - 3y
13find the lapican of the function. f(x,y) = x^2 + y^2
- A2 + y^2 + x^2 + 2
- B2x + 2y + 2x + 2y
14what is the correct matrix for this function f1(x,y) = x^2 + y^2 and f2(x,y) = x^3 + y^3
- A
- B
- A
15what is the matrix of the jacobian when x = 1 and y = 1
- A
- B
- A
16what is the value of change in the matrix?
- A44
- B20
Información
- Descripción
- (week 12) multaviarible calculus Test sobre Unit 2 test, creado por jason wheeler el 16/10/2017.
- Autor
- jason wheeler
- Idioma
- English
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- 0
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