Trouble with Tangent of a Line

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To find the equation of the line perpendicular to the tangent of y=5x^2 at the point (1,5), the slope of the tangent was calculated to be approximately 10. The negative reciprocal of this slope is -1/10, which is used to determine the slope of the perpendicular line. The equation of the line can be derived using the point-slope form, resulting in x + 10y - 51 = 0. The confusion arose regarding the constant 51 in the equation, which was clarified through substitution. Overall, the discussion emphasizes the importance of understanding slopes and their relationships in determining perpendicular lines.
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 y=5x^2 at (1,5)

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

(5x^2-5)/(x-1)

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 y=5x^2 at (1,5)

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

(5x^2-5)/(x-1)

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 y - y_{1} = -\frac{1}{10}(x-x_{1}), where (x_{1}, y_{1})=(1, 5). 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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