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Calculus/Normal Line To Curve

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Expert: - 4/4/2009


Question
The equation of the normal line to the curve
y = (x^2 - 1)^(1/3) at the point where x = 3 is

a. y + 12x = 38
b. y - 4x = 10
c. y + 2x = 4
d. y + 2x = 8
e. y - 2x = -4

Answer

Line and curve near x=
To find this, the equation must have the same slope.
To find the answer, take the derivative and put in 3.
This will give the slope and you can then do
(y-y3) = m(x-3) where y3 is the value of y at x=3.

The value at 3 should not be hard to find.
Take 3 and put it in the equation (x^2 - 1)^(1/3),
and you get (3^2-1)^(1/3) = (9-1)^(1/3) = 8^(1/3) = 2.

There are two approaches: finish the problem or try the answers.

We know that x=3 and y=2, so
a) y + 12x = 38 => 2 + 12*3 = 38?  Yes, that could be it.

b) y - 4x = 10 => 3 - 4*2 = 10?  
No, the ansswer should be -5, not 10, so that's not it.

c) 3y + 2x = 4 => 3(2) + 2(3) = 4?  
No, the answer should be 12, not 4, so that's not it.

d) y + 2x = 8 => 2 + 3(3) = 8?  
No, the answer is 11 and it should be 8, so that doesn't work.

e) y - 2x = 4 => 2 - 2(3) = 4?
No, the answer is -6, and it should be 4, so that doesn't work.

The answer looks like (a), but just to be sure, let's see what it is.
The derivative is a chain rule.

It can be seen that y' = (1/3)(x²-1)^(1/3 - 1)(2x).
The 1/3 -1 = -2/3, so the exponet is 2/3 in the denominator.

This says that y' = 2x/[3(x²-1)^(2/3)].  
The slope at x=3 is then 2(3)/[3(3²-1)^(2/3)] = 6/[3(8)^(2/3)]
= 6/(3*2²) = 6/12 = 1/2.

At that, (a) is not correct either.

I have included a view of the graph and the line to show that the
answer should be f. none of the above.

Scotto

Expertise

Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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Experience in the area: I have tutored students in all areas of mathematics for over 25 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

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