Equation of Normal to Curve at (1,5): Solved?

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Discussion Overview

The discussion revolves around finding the equation of the normal line to the curve defined by the equation y = x² + 4 at the point (1, 5). Participants explore the derivation of the tangent line and the subsequent normal line, addressing potential errors in calculations and assumptions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about finding the equation of the normal line at the point (1, 5) on the curve y = x² + 4.
  • Another participant confirms the slope of the tangent line at (1, 5) is 2, derived from the derivative y' = 2x.
  • There is a discussion about the tangent line equation being y = 2x + 3, with a participant emphasizing that the normal line, not the tangent, is the focus of the problem.
  • One participant points out that the point (1, 5) does not satisfy the equation x + 2y = 10, suggesting a need to adjust the constant in that equation.
  • A later reply provides the equation of the normal line as y = -x/2 + 11/2, based on the established slope of the tangent line.

Areas of Agreement / Disagreement

Participants generally agree on the slope of the tangent line being 2, but there is disagreement regarding the correctness of the initial equation proposed for the normal line and the calculations leading to it. The discussion remains unresolved regarding the final form of the normal line equation.

Contextual Notes

There are indications of confusion regarding the substitution of values into equations and the distinction between the tangent and normal lines. Some assumptions about the correctness of initial equations are challenged without resolution.

alexandravo
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Really confused bout a question and finding the equation.

A normal is drawn at the point (1,5) on the curve defined by the rule y=x2+4. Find the equation of the normal.

I substituted the values x=1 and y=5 into the derived equation and got my answer to be x+2y=10? Is that correct?
 
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alexandravo said:
Really confused bout a question and finding the equation.

A normal is drawn at the point (1,5) on the curve defined by the rule y=x2+4. Find the equation of the normal.

I substituted the values x=1 and y=5 into the derived equation and got my answer to be x+2y=10? Is that correct?
You got the slope right. But the point $(1,5)$ does not lie on the line $x+2y = 10$, so you need to adjust the constant $10$.
 
If y=x^2+ 4, y'= 2x. (1, 5) is on the curve so the tangent line through (1, 5) has slope 2(1)= 2.
The tangent line is y(x)= 2x+ C where C is such that y(1)= 2(1)+ C= 5. C= 5- 2= 3. The tangent line is y= 2x+ 3.

You say "I substituted the values x=1 and y=5 into the derived equation". I assume you mean the equation y'= 2x. Yes, with x= 1, y', the slope, is 2 but there is no reason to calculate 2 times 5! y is 5, not x! And y' is the slope of the tangent line, not the constant term.
 
Country Boy said:
If y=x^2+ 4, y'= 2x. (1, 5) is on the curve so the tangent line through (1, 5) has slope 2(1)= 2.
The tangent line is y(x)= 2x+ C where C is such that y(1)= 2(1)+ C= 5. C= 5- 2= 3. The tangent line is y= 2x+ 3.

You say "I substituted the values x=1 and y=5 into the derived equation". I assume you mean the equation y'= 2x. Yes, with x= 1, y', the slope, is 2 but there is no reason to calculate 2 times 5! y is 5, not x! And y' is the slope of the tangent line, not the constant term.
But we're looking for the normal, not the tangent!
 
Opalg said:
You got the slope right. But the point $(1,5)$ does not lie on the line $x+2y = 10$, so you need to adjust the constant $10$.
Country Boy Is correct in stating that the slope of the tangent line is 2. So the tangent line equation is
y = 2x + 3
and the equation of the normal is then
y = -x/2 + 11/2
 

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