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What is the equation of the line normal to # f(x)=e^(sqrtx)# at # x=4#?

Answer 1

Christopher Agresta

It is #y=-4/e^2x+9.55#.

The line passing from #x=4# and tangent to #e^sqrt(x)# has as slope

#d/dxe^sqrt(x)# evaluated when #x=4#. The derivative can be calculated with the chain's rule

#d/dxe^sqrt(x)=e^sqrt(x)*d/dxsqrt(x)=e^sqrt(x)/(2sqrt(x))#.

The slope is

#m=e^sqrt(4)/(2sqrt(4))=e^2/4#.

So the tangent has equation

#y=e^2/4x+q#

The #q# can be found imposing the passage of the line from the point #(4, e^sqrt(4))=(4, e^2)#

#e^2=e^2/4*4+q# then #q=0# and the equation of the tangent is

#y=e^2/4x#

The orthogonal has the negative inverse slope, so the equation is

#y=-4/e^2x+q#

Again, to find #q# we have to impose the passage from #(4, e^2)#

#e^2=-4/e^2*4+q#

#q=e^2+16/e^2=(e^4+16)/e^2\approx9.55# and finally the equation of the orthogonal is

#y=-4/e^2x+9.55# or

#y=-0.54x+9.55#.

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Answer 2

Luke Alvarez

The equation of the line normal to f(x) = e^(sqrt(x)) at x = 4 is y = -2e^(2) * (x - 4) + e^(4).

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