What is the slope of the line tangent to the curve y+2=(x^2/2)-2sin(y) at the point (2,0)?
2019-01-03 10:59 am
I don’t even know where to start so if you could show all work or explain that would be great
回答 (4)
2019-01-03 11:05 am
dy/dx = x - 2cos(y) dy/dx
dy/dx(1 + 2cos(y)) = x
dy/dx = x/(1 + 2cos(y))
at (2,0), m = dy/dx = 2/(1 + 2cos(0)) = 2/(1 + 2) = 2/3
dy/dx(1 + 2cos(y)) = x
dy/dx = x/(1 + 2cos(y))
at (2,0), m = dy/dx = 2/(1 + 2cos(0)) = 2/(1 + 2) = 2/3
2019-01-03 11:03 am
Derive implicitly
y + 2 = (1/2) * x^2 - 2 * sin(y)
dy + 0 = (1/2) * 2x * dx - 2 * cos(y) * dy
dy = x * dx - 2 * cos(y) * dy
x = 2 , y = 0
dy = 2 * dx - 2 * cos(0) * dy
dy = 2 * dx - 2 * 1 * dy
dy = 2 * dx - 2 * dy
dy + 2 * dy = 2 * dx
3 * dy = 2 * dx
3 * dy/dx = 2
dy/dx = 2/3
y + 2 = (1/2) * x^2 - 2 * sin(y)
dy + 0 = (1/2) * 2x * dx - 2 * cos(y) * dy
dy = x * dx - 2 * cos(y) * dy
x = 2 , y = 0
dy = 2 * dx - 2 * cos(0) * dy
dy = 2 * dx - 2 * 1 * dy
dy = 2 * dx - 2 * dy
dy + 2 * dy = 2 * dx
3 * dy = 2 * dx
3 * dy/dx = 2
dy/dx = 2/3
2019-01-03 4:33 pm
y + 2 = (x²/2)-2siny at point (2, 0)
dy/dx + 0 = 2x/2 - 2cosy.dy/dx
dy/dx = x-2cosy dy/dx
dy/dx + 2cosy.dy/dx = x
dy/dx (1+2cosy) = x
dy/dx = (x/1 + 2cosy) / (2, 0)
= 2/1 + 2cos0 = 2/3.
dy/dx + 0 = 2x/2 - 2cosy.dy/dx
dy/dx = x-2cosy dy/dx
dy/dx + 2cosy.dy/dx = x
dy/dx (1+2cosy) = x
dy/dx = (x/1 + 2cosy) / (2, 0)
= 2/1 + 2cos0 = 2/3.
2019-01-03 12:40 pm
take a differential
dy = 2xdx / 2 - 2 cos(y) dy
dy / dx = x / ( 1 + 2cos(y))
dy / dx = 2 / ( 1 + 2cos(0))
dy / dx = 2/3
dy = 2xdx / 2 - 2 cos(y) dy
dy / dx = x / ( 1 + 2cos(y))
dy / dx = 2 / ( 1 + 2cos(0))
dy / dx = 2/3
收錄日期: 2021-05-01 20:52:55
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