Trouble with Tangent of a Line

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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.
 
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?
 
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.
 
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.)
 
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
 
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: