Question: The slope of any point (x,y) of a curve is given by dy/dx = (√x ) (4 - x). If the curve passes through the point (1,1), find the equation of the curve.
With detailed steps please.
A.maths question about integration
2007-10-19 11:46 pm
回答 (3)
2007-10-20 12:14 am
✔ 最佳答案
dy/dx = (√x ) (4 - x)= 4x^(1/2) - x^(3/2)
y = int [ 4x^(1/2) - x^(3/2) ] dx
= (8/3)x^(3/2) - (2/5)x^(5/2) + C, C is constant
= (8/3)x^(3/2) - (2/5)x^(5/2) + C
(8/3)(1)^(3/2) - (2/5)(1)^(5/2) + C = 1
(8/3) - (2/5) + C = 1
C = -19/15
y = (8/3)x^(3/2) - (2/5)x^(5/2) - 19/15
2007-10-20 12:16 am
dy/dx = (√x ) (4 - x)
dy/dx = 4x^(1/2) - x^(3/2)
y = ∫(4x^(1/2) - x^(3/2)) dx
y = 4x^(3/2)/(3/2) - x^(5/2)/(5/2) + C
y = 8x^(3/2)/3 - 2x^(5/2)/5 + C
at (1, 1)
1 = 8/3 - 2/5 + C
C = -19/15
equation of the curve is y = y = 8x^(3/2)/3 - 2x^(5/2)/5 - 19/15
dy/dx = 4x^(1/2) - x^(3/2)
y = ∫(4x^(1/2) - x^(3/2)) dx
y = 4x^(3/2)/(3/2) - x^(5/2)/(5/2) + C
y = 8x^(3/2)/3 - 2x^(5/2)/5 + C
at (1, 1)
1 = 8/3 - 2/5 + C
C = -19/15
equation of the curve is y = y = 8x^(3/2)/3 - 2x^(5/2)/5 - 19/15
2007-10-20 12:08 am
X*3.14/4=curve
收錄日期: 2021-04-13 14:47:35
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