Trouble with Tangent of a Line

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Homework Help Overview

The discussion revolves around determining the equation of a line that is perpendicular to the tangent of the function y=5x^2 at the point (1,5). Participants explore the concepts of slopes and tangents in the context of calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the formula for instantaneous rate of change and the slope of the tangent line. There are attempts to differentiate the function to find the slope, and questions about the relationship between the slope of a line and the slope of a line perpendicular to it are raised.

Discussion Status

Some participants have provided guidance on finding the slope of the perpendicular line and the equation format. There is a recognition of the negative reciprocal relationship between the slopes, and some confusion about specific calculations has been noted, but no consensus on the final equation has been reached.

Contextual Notes

Participants express uncertainty about specific values in the equation derived and the implications of their calculations. There is an acknowledgment of potential misunderstandings in the process of deriving the equation.

thomasrules
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I tried many thigns with this one:

Determine the equation of the line that is perpendicular to the tangent to [tex]y=5x^2 at (1,5)[/tex]

What I did was use the formula for instantaneous rate of change finding the equation of the secant.

[tex](5x^2-5)/(x-1)[/tex]

from there I found that the slope of the tangent is about 10 and from there I'm lost
 
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thomasrules said:
I tried many thigns with this one:

Determine the equation of the line that is perpendicular to the tangent to [tex]y=5x^2 at (1,5)[/tex]

What I did was use the formula for instantaneous rate of change finding the equation of the secant.

[tex](5x^2-5)/(x-1)[/tex]

from there I found that the slope of the tangent is about 10 and from there I'm lost

Maybe it would be easier to differentiate the function and plug in the x-coordinate to find the slope of the tangent.
 
but I have the slope...
 
thomasrules said:
but I have the slope...

Ok then. In what relation are the slope of a line with the slope of a line perpendicular to it?
 
duh negative reciprocal
 
thomasrules said:
duh negative reciprocal

So, all you have to do now is find the equation of the line which passes through the given point and whose slope is, as you stated, negative reciprocal to the slope of the tangent at that very same point.
 
answer is x+10y-51=0 but whres the 51 from
 
thomasrules said:
answer is x+10y-51=0 but whres the 51 from

The equation of the line is [tex]y - y_{1} = -\frac{1}{10}(x-x_{1})[/tex], where [tex](x_{1}, y_{1})=(1, 5)[/tex]. Plug in the values and you should get the equation. (And see where 51 comes from.)
 
ur not suggesting y-y1=m(x-x1)

??
 
  • #10
thats what i did one sec
 
  • #11
omg I'm so smart because I got to that last step u said but didnt see that 1/10 +5 is 5.1...

but that's why I'm so stupid too i didnt see it
 
  • #12
thomasrules said:
omg I'm so smart because I got to that last step u said but didnt see that 1/10 +5 is 5.1...

but that's why I'm so stupid too i didnt see it

You're not stupid, you're just slumpy. That can be cured easily. :smile:
 
  • #13
Thank you Mr Radou for your help and compliment
 

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