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secondary 4 | A Maths
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Date Posted: 6 years ago
Views: 753
You are given the double derivative of the equation:
D2y/dx2 =-5
If you integrate it you will get
Dy/dx = -5x+C
You are given the info at that point p (-1,3),
Gradient =2
So, substituting the info you have:
Dy/dx=2, x=-1
2=-1 (-5)+C
2=5+C
C= -3
So,
Dy/dx = -5x -3
Integrating the equation again you get
Y= -(5x^2)/2 -3x + C
Sub point P, y=3, x= -1
Solve for C, and then you will get the whole equation.
Hope it helped!
D2y/dx2 =-5
If you integrate it you will get
Dy/dx = -5x+C
You are given the info at that point p (-1,3),
Gradient =2
So, substituting the info you have:
Dy/dx=2, x=-1
2=-1 (-5)+C
2=5+C
C= -3
So,
Dy/dx = -5x -3
Integrating the equation again you get
Y= -(5x^2)/2 -3x + C
Sub point P, y=3, x= -1
Solve for C, and then you will get the whole equation.
Hope it helped!
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Date Posted:
6 years ago
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