# Find the slope of the curve tan^-1(2x/y)=(πx/(y^2)) at the point (1, 2)π=pi

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember that the slope of the curve, represented by a function, at a point, is evaluated using derivative of the function at the point.

Hence, you should differentiate both sides with respect to x such that:

`(d(tan^-1(2x/y)))/(dx) = (d(pix/(y^2)))/(dx)`

`1/(1 + 4x^2/y^2)*(d(2x/y))/(dx) = (pi*y^2 - pi*x*2y*(dy)/(dx))/(y^4)`

`1/(1 + 4x^2/y^2)*(2y - 2x(dy)/(dx))/y^2 = (pi*y^2 - pi*x*2y*(dy)/(dx))/(y^4)`

`(2y - 2x(dy)/(dx))/(y^2 + 4x^2) = (pi*y^2 - pi*x*2y*(dy)/(dx))/(y^4)`

`y^4(2y - 2x(dy)/(dx)) = (pi*y^2 - pi*x*2y*(dy)/(dx))(y^2 + 4x^2)`

You need to open the brackets such that:

`2y^5 - 2xy^4(dy)/(dx) = pi*y^4 + 4pix^2*y^2 - 2pi*x*y^3*(dy)/(dx) - 8pix^3*y(dy)/(dx)`

You need to isolate the terms that contain `(dy)/(dx)`  to the left side such that:

`2pi*x*y^3*(dy)/(dx)+ 8pix^3*y(dy)/(dx) - 2xy^4(dy)/(dx) = pi*y^4 + 4pix^2*y^2 - 2y^5`

Factoring out `(dy)/(dx)`  yields:

`(dy)/(dx)(2pi*x*y^3 + 8pix^3*y - 2xy^4) = pi*y^4 + 4pix^2*y^2 - 2y^5`

You need to divide by `(2pi*x*y^3 + 8pix^3*y - 2xy^4)`  such that:

`(dy)/(dx) = (pi*y^4 + 4pix^2*y^2 - 2y^5)/(2pi*x*y^3 + 8pix^3*y - 2xy^4)`

You need to evaluate the slope `(dy)/(dx)`  at the point (1,2) such that:

`(dy)/(dx)|_(1,2) = (pi*2^4 + 4pi*1^2*2^2 - 2^5)/(2pi*1*2^3 + 8pi*1^3*2 - 2*1*2^4)`

`(dy)/(dx)|_(1,2) = (16pi + 16pi - 32)/(16pi + 16pi - 32)`

`(dy)/(dx)|_(1,2) = (32 pi - 32)/(32 pi - 32)`

`(dy)/(dx)|_(1,2) = 1`

Hence, evaluating the slope of the tangent line to the given curve, at (1,2), yields `(dy)/(dx)|_(1,2) = 1` .