SOLUTION: Hi can somebody please help with the below, and be as kind as to talk me through the steps, i am totally confused with this.
Find dy=dx as a function of x and y when:
i) x^2
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Quadratic Equations and Parabolas
-> SOLUTION: Hi can somebody please help with the below, and be as kind as to talk me through the steps, i am totally confused with this.
Find dy=dx as a function of x and y when:
i) x^2
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Question 376103: Hi can somebody please help with the below, and be as kind as to talk me through the steps, i am totally confused with this.
Find dy=dx as a function of x and y when:
i) x^2y^3 - x^2 + 3y - 3 = 0
ii)(1 - x^2) tan^-1 y = 3
Really appreciate this.
Thanks
You can put this solution on YOUR website! Find dy/dx as a function of x and y when:
i) x^2y^3 - x^2 + 3y - 3 = 0
Differentiate implicitly
2x*dx*y^3 +x^2*3y^2*dy - 2xdx + 3dy = 0
dx*(2xy^3 - 2x) + dy*(x^2*3y^2 + 3) = 0
dy*(3x^2y^2 + 3) = -dx(2xy^3 - 2x)
dy/dx = -2(xy^3 - x)/(3x^2y^2 + 3)
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ii)(1 - x^2) tan^-1 y = 3
Use atan for arctan, easier to type
atan(y) - x^2*atan(y) = 3
dy/(1+y^2) - (2x*dx*atan(y) + x^2*dy/(1+y^2) = 0
Multiply thru by (1+y^2)
dy - 2x*dx*atan(y)*(1+y^2) - x^2*dy = 0
(1-x^2)dy = 2x*atan(y)*(1+y^2)dx
dy/dx = 2x*atan(y)*(1+y^2)/(1-x^2) with y <> ± 1
