For what non-negative value of b is the line given by y=-1/3x+b normal to the curve y

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davidiswhat

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For what non-negative value of b is the line given by y=-1/3x+b normal to the curve y=x^3?
so my thought process
y'= -1/3 + 0 is the slope of the tangent of y=x^3
y'=3x^2 is the slope of y=x^3
this is where im confuse why is it equal to 3
y'=3x^2 =3 due to that being a perpendicular slope to -1/3? i dont think the tangent slope has to be perpendicular ...
3x^2 = 3
x^2=1
x=+-1
y(1) = (1)^3 =(1,1) is the point where y= x^3 has a slope/normal of -1/3
x=1 because when we plug it back in y=-1/3x+b b = a positive when we solve for x
b=4/3
 
Who ever gave you this problem expects you to know that "lines y= mx+ b and y= nx+ a are perpendicular if and only if mn= -1". Yes, the given line has slope -1/3. So you want the slope of the tangent line, m, to satisfy m(-1/3)= -1. Solve that for m.
 
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No, it is not. The line you are given has slope -1/3 and you want the tangent line to be perpendicular to it, not parallel.
 
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