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Find the equation of the normal line of the given parabola that is parallel to a line

  • Thread starter illjazz
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  • #1
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Homework Statement


Find an equation of the normal line to the parabola y = x^2 - 5x + 4 that is parallel to the line x - 3y = 5.


Homework Equations


? All that is relevant is mentioned below AFAIK.


The Attempt at a Solution


Here is my attempt: http://screencast.com/t/oXxI6JPHv

My book says the solution is supposed to be
y = 1/3x - 1/3

I also found this: http://answers.yahoo.com/question/i...L8aRlYEjzKIX;_ylv=3?qid=20071125140023AAebnO1

It's the exact same problem and algebraically, I arrive at the same results as the person who answered the question there.

I also don't see that the negative reciprocal of the slope was taken anywhere. After all, I need to find the equation of the normal line to the parabola and not the equation of the tangent line. What am I missing?
 
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Answers and Replies

  • #2
rock.freak667
Homework Helper
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The gradient of the normal is 1/3. So the gradient of the tangent to the parabola is **?
And you want to find the point where the gradient of the tangent is ** i.e. y/=**
 
  • #3
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The gradient of the normal is 1/3. So the gradient of the tangent to the parabola is **?
And you want to find the point where the gradient of the tangent is ** i.e. y/=**
Eh? Gradient? This problem is from section 3.1 in Calculus Concepts & Contexts, 3rd ed. by James Stewart, and I just double-checked.. there is no mention of the term "gradient" anywhere in there.

I don't doubt that what you've suggested is correct--I'm just saying that I'm not familiar with the concept of the gradient at all.
 
  • #4
rock.freak667
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Gradient is the same as slope.
 
  • #5
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Gradient is the same as slope.
Whoops! Thank you.

So, if the slope of the normal line is 1/3 and the slope of the normal line is the negative reciprocal of the slope of the tangent line, I assume I can apply the same idea in reverse? In other words, if the slope of the normal is 1/3, would the slope of the tangent then be -3?

Taking the first derivative of the original function, I thought, gives the equation for the rate of change of that function.. i.e., the equation for the tangent line at a given point. What I find a bit confusing is what meaning the derivative of a function has when it is taken WITHOUT a point on the curve. I guess it just means the general formula one can use to find the slope at a specific point when the coordinates of this point are given, correct?

Further, it appears that I don't really have to bother thinking about the tangent here at all, since it looks like we are skipping to finding the normal line's equation directly.

What's bothering me is that I don't see what I am missing. My calculations are correct as far as I can tell and yet I still do not end up at 1/3x - 1/3, which is what my book shows as the solution to this problem.
 
  • #6
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What # is this in your book? I'm getting the same answer as the yahoo post. If you graph the solution your book gives you, it doesn't even come close to touching your parabola.
 
  • #7
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What # is this in your book? I'm getting the same answer as the yahoo post. If you graph the solution your book gives you, it doesn't even come close to touching your parabola.
It's #51 in section 3.1, (exercises start on p. 190).
 
  • #8
rock.freak667
Homework Helper
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So, if the slope of the normal line is 1/3 and the slope of the normal line is the negative reciprocal of the slope of the tangent line, I assume I can apply the same idea in reverse? In other words, if the slope of the normal is 1/3, would the slope of the tangent then be -3?
Yes the slope of the tangent at that point (x,y) is -3


Taking the first derivative of the original function, I thought, gives the equation for the rate of change of that function.. i.e., the equation for the tangent line at a given point. What I find a bit confusing is what meaning the derivative of a function has when it is taken WITHOUT a point on the curve. I guess it just means the general formula one can use to find the slope at a specific point when the coordinates of this point are given, correct?
Yes that is correct.Your first derivative is the gradient function, so you can find the gradient of the tangent to the curve at any point onthe curve.

Further, it appears that I don't really have to bother thinking about the tangent here at all, since it looks like we are skipping to finding the normal line's equation directly.
You don't need to bother knowing the equation of the tangent really.

To find the equation of any line, you need:
1. The gradient of the line (Which you have)
2. A point which is on the line (Which you need to find)

What's bothering me is that I don't see what I am missing. My calculations are correct as far as I can tell and yet I still do not end up at 1/3x - 1/3, which is what my book shows as the solution to this problem.

y = x^2 - 5x + 4

[tex]\frac{dy}{dx}=2x-5[/tex]


You agree that that this gives the gradient of the tangent at any point (x,y) on the curve right?

(x,y) is the point you want to find right?

If the gradient of the tangent at (x,y) is -3. Doesn't that mean, at that point [itex]\frac{dy}{dx}=-3[/itex] ?

You can now find the x-coordinate of the point that you want.
 
  • #9
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Yes the slope of the tangent at that point (x,y) is -3

Yes that is correct.Your first derivative is the gradient function, so you can find the gradient of the tangent to the curve at any point onthe curve.
Thanks! Knowing that I have the right idea here helps a lot.

You don't need to bother knowing the equation of the tangent really.

To find the equation of any line, you need:
1. The gradient of the line (Which you have)
2. A point which is on the line (Which you need to find)
Again, much appreciated. It's nice to see "absolute" rules from time to time--ones that can be relied on no matter what :)

y = x^2 - 5x + 4

[tex]\frac{dy}{dx}=2x-5[/tex]


You agree that that this gives the gradient of the tangent at any point (x,y) on the curve right?

(x,y) is the point you want to find right?

If the gradient of the tangent at (x,y) is -3. Doesn't that mean, at that point [itex]\frac{dy}{dx}=-3[/itex] ?

You can now find the x-coordinate of the point that you want.
Ooooh.. when I read that I felt an "a-ha!" moment creeping up! Thanks! I will have to get back to this problem a little later in the day but I already see where this is going. I'll post back with the results, of course!
 
  • #10
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Success! Thank you so much. I even understand why it ended up being this way.. basically, that Yahoo answer threw me off completely because the guy who answered that ignored the "normal line" requirement entirely. He appears to just have found the tangent line.. not the normal line.

Anyway. I ended up with this:

[tex]y'=2x-5[/tex]
[tex]2x-5=-3[/tex]
[tex]x=1[/tex]


Plugging that back into y gave:

[tex]y=1^2-5(1)+4=0[/tex]


Which gives me the point (1, 0).

The equation for the normal line then becomes:

[tex]y-0=\frac{1}{3}(x-1)[/tex]
[tex]y=\frac{1}{3}x-\frac{1}{3}[/tex]

Which is exactly what my book says the solution is!

Again, thank you!

By the way.. why does "\\" not produce a new line/linebreak when using Latex? I looked at the crash course/reference and on the web as well.. no matter how I tried, \\ will NOT give me a new line :/
 
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