Find the equation of the normal to the curve y= x(1-2x)^{1/2 }at the point (-4,-12)

y= x(1-2x)^{1/2 }dy/dx = 1/2 (1-2x)^{-1/2}

do i sub in -4 now? If so, am i getting the gradient of the normal or the curve itself?

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- Mar 30th 2013, 07:26 PMThorpeliztsDifferenciation
Find the equation of the normal to the curve y= x(1-2x)

^{1/2 }at the point (-4,-12)

y= x(1-2x)^{1/2 }dy/dx = 1/2 (1-2x)^{-1/2}

do i sub in -4 now? If so, am i getting the gradient of the normal or the curve itself? - Mar 30th 2013, 10:10 PMibduttRe: Differenciation
Your derivation is wrong. Use product rule

d(uv)/dx = u dv/dx + v du/dx.

the derivative comes out to be dy / dx = (1-3x)/(1-2x)^1/2

Now get dy/dx at x = -4

that is the slope of tangent at x = -4.

Negative reciprocal of this slope will give us slope of normal to the curve at x = -4.

Thus slope of normal would be -1/[dy/dx] = - (1-2x)^1/2 / (1-3x)

Now you can proceed further