Posted by: comingsoon February 28, 2011
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1) A normal line to the graph of a function f at the point (x,f(x)) is defined to be the line perpendicular to the tangent line at that point . Find the equation of the normal line to the curve y= ³√x² at the point where x=3
Solution:
the point is (3, 3 2/3).................. [³√3²=3 2/3]
Now for slope m,
dy/dx=d(x2/3)/dx= 2/3 x-1/3
at poin x=3
2* 3 -1/3/3= 2/(3* 3 1/3) = 2/3 4/3
The line is perpendicular is m=-3 4/3/2
finally the equation of the normal line is
y-32/3/2=3 4/3/2(x-3)
y=(-3 4/3/2 )x +3 7/2 +3 2/3
Solution:
the point is (3, 3 2/3).................. [³√3²=3 2/3]
Now for slope m,
dy/dx=d(x2/3)/dx= 2/3 x-1/3
at poin x=3
2* 3 -1/3/3= 2/(3* 3 1/3) = 2/3 4/3
The line is perpendicular is m=-3 4/3/2
finally the equation of the normal line is
y-32/3/2=3 4/3/2(x-3)
y=(-3 4/3/2 )x +3 7/2 +3 2/3