Find equation of normal line to y = x^1/2 that is parallel to y = 1 - 2x

[ricecrispie]

Completely stumped by this question:

Q:

Find the equation of the normal line to the curve y = x^1/2 that is parallel to the line y = 1 - 2x

So we know:

m is -2 as the normal line is // to 1-2x
y' = 1/2(x^-1/2)

The normal line is perpendicular to the tangent line, therefore m * y' = -1, therefore the gradient of the normal line must be 1/2.

So then:

1/2 = y' = 1/2(x^-1/2)
1 = x^-1/2
Therefore x = 1

y(1) = 1 ....... (1,1)

y - 1 = 1/2(x - 1)
y = x/2 + 1/2

Required answer : y = -2x + 3

I don't think I'm understanding what the question is asking, I also tried using m = -2

Alternative answer:

-2 = y' = 1/2(x^-1/2)
-4 = x^-1/2
But that's impossible right?

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[Dr.Peterson]

Read your question as if someone else sent it to you, and you want to check it out. First look at the problem:

Find the equation of the normal line to the curve y = x^1/2 that is parallel to the line y = 1 - 2x

So the answer has to be the equation of a line whose slope is -2, right?

What is the slope of your answer?

Can you see where you got that from, and what mistake led to it?

Hint: You are saying something about the normal line that you should have said about the tangent ... Most of your work is correct, but you applied it incorrectly.

 

[ricecrispie]

Read your question as if someone else sent it to you, and you want to check it out. First look at the problem:



So the answer has to be the equation of a line whose slope is -2, right?

What is the slope of your answer?

Can you see where you got that from, and what mistake led to it?

Hint: You are saying something about the normal line that you should have said about the tangent ... Most of your work is correct, but you applied it incorrectly.

I see that the gradient of the normal line must be -2 as the question wants it to be parallel to y = 1 -2x but since the normal line is perpendicular to the tangent line, y' should equal 1/2 ?

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[Dr.Peterson]

I see that the gradient of the normal line must be -2 as the question wants it to be parallel to y = 1 -2x but since the normal line is perpendicular to the tangent line, y' should equal 1/2 ?

Correct. So when you said, "therefore the gradient of the normal line must be 1/2", you meant "of the tangent", right?

But then you didn't write the equation of the normal line! Your equation is the tangent.

 

[ricecrispie]

Correct. So when you said, "therefore the gradient of the normal line must be 1/2", you meant "of the tangent", right?

But then you didn't write the equation of the normal line! Your equation is the tangent.

Oh yes I understand ! But we're still looking for the equation of the normal line not tangent so what do I do with the fact that the slope of the tangent line is 1/2? I get confused by that

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[Dr.Peterson]

Oh yes I understand ! But we're still looking for the equation of the normal line not tangent so what do I do with the fact that the slope of the tangent line is 1/2? I get confused by that

Who cares what the slope of the tangent is, when you are asked for the normal? Just use the slope of the normal in the equation:

y - 1 = -2(x - 1)

and so on.

 

[sinx]

Oh yes I understand ! But we're still looking for the equation of the normal line not tangent so what do I do with the fact that the slope of the tangent line is 1/2? I get confused by that

Sent from my LG-H840 using Tapatalk

I believe you are thinking about it correctly.
when you know the slope of the tangent, the normal line will be 90⁰ to that, and is the negative reciprocal.
but prove it to yourself. Then it will sink in.
draw a line with slope of 1/2. You should find that a slope of -2 is at 90⁰ to it;

{and 1/3 is 90⁰ to slope=-3, and so on.}