Sun K.

asked • 12/29/13

For what non-negative value?

For what non-negative value of b is the line given by y=-1/3x+b normal to the curve y=x^3?
 
Answer: 4/3
 
 

Morgan O.

Y=x3
Y'=3x2
Normal slope=  -1/3
Y= -1/33=3
Y'=(3x2=3)/3=(√x2=1)
X=1
(1,1)
Y=(-1/3)x+b
1=(-1/3)(1)+b
1=(-1/3)+b
b=4/3
 
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10/28/14

1 Expert Answer

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Inactive Tutor answered • 12/29/13

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y = x^3
y' = 3x^2 = 3, since its normal line has a slope of -1/3.
Solve for x,
x = 1, y = x^3 = 1
Plug in (1, 1) into y = (-1/3)x + b,
1 = -1/3 + b
b = 4/3 <==Answer
-------
Attn: x = -1 leads to negative value of b.
 
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Sun K.

But how did you get x=1?
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12/30/13

Sun K.

Never mind. I got it.
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12/30/13

Kervin M.

Why did you set the derivative equal to 3?
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09/26/22

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