Finding the equation of the normal line at

In summary, the conversation discussed finding the equation of the normal line to the graph of y=3x^4 - 1/x at x=1. The derivative of 3x^4 was correctly found to be 12x^3, and the derivative of 1/x was corrected to be 1/x^2. The slope of the normal line was found to be -1/13, and the final equation was derived to be x + 13y - 27 = 0. A minor mistake was acknowledged, but it did not affect the outcome.
  • #1
meeklobraca
189
0

Homework Statement



Find the equation, in standard form, of the normal line drawn to the graph of y=3x4 - 1/x at x = 1



Homework Equations





The Attempt at a Solution



-I figured out the slope of the tangent to be 12x3 + 1, then at x = 1 the slope is 13
-SO the slope of the normal being the negative reciprocal of 13 led me to get -1/13
-Plug that into the y - y1 = m(x - x1) equation and I get x +13y - 27 = 0 for the equation of the normal.

Im a little gun shy since the numbers seem abnormally large, so I would love if someone could verify this, or if I am wrong, point me in the right direction of being right!

Cheers.
 
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  • #2
meeklobraca said:

Homework Statement



Find the equation, in standard form, of the normal line drawn to the graph of y=3x4 - 1/x at x = 1



Homework Equations





The Attempt at a Solution



-I figured out the slope of the tangent to be 12x3 + 1, then at x = 1 the slope is 13
-SO the slope of the normal being the negative reciprocal of 13 led me to get -1/13
-Plug that into the y - y1 = m(x - x1) equation and I get x +13y - 27 = 0 for the equation of the normal.

Im a little gun shy since the numbers seem abnormally large, so I would love if someone could verify this, or if I am wrong, point me in the right direction of being right!

Cheers.

You are right that the derivative of [tex]3 {x^4}[/tex] is [tex]12{x^3}[/tex]

But what is the derivative ot 1/x again?
 
  • #3
You made a minor mistake in your derivative, it should be 12x³ + 1/x², not 12x³ + 1. However, this doesn't affect the slope at x=1, and the rest of your derivation (including the answer) is correct.
 
  • #4
YUP i see where I made the mistake. Instead of subtracting the 1 from the x-1 I added. But your right it doesn't affect the outcome. But thanks guys!
 

What is the normal line?

The normal line is a line that is perpendicular to a given curve or surface at a specific point. It intersects the curve or surface at a 90-degree angle and represents the slope of the curve or surface at that point.

How do you find the equation of the normal line at a given point?

To find the equation of the normal line at a given point, you need to first find the slope of the tangent line at that point. Then, you can use the negative reciprocal of the tangent slope to find the slope of the normal line. Finally, you can use the point-slope form of a line to write the equation of the normal line.

Can the normal line exist at any point on a curve or surface?

Yes, the normal line can exist at any point on a curve or surface, as long as the curve or surface has a defined slope at that point. However, some curves or surfaces may have multiple normal lines at a single point.

Is the normal line the same as the tangent line?

No, the normal line and the tangent line are not the same. The tangent line represents the instantaneous rate of change, or slope, of a curve or surface at a specific point, while the normal line represents the slope that is perpendicular to the tangent line at that point.

Why is the normal line important in mathematics?

The normal line is important in mathematics because it helps us understand the behavior of curves and surfaces at specific points. It also allows us to find the slope of these curves and surfaces at any given point, which is crucial in many mathematical applications.

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