normal line to the curve

Oct 2009
202
4
A curve \(\displaystyle y=f(x)\) with the gradient function \(\displaystyle \frac{dy}{dx}=\frac{x^2}{3x-1}.\)

The straight line \(\displaystyle ky=2x-7\) is normal to the curve \(\displaystyle y=f(x)\) at \(\displaystyle (2,m)\).Find the value of \(\displaystyle k\) and \(\displaystyle m\).


any help will appreciate,.
 
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mr fantastic

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The straight line \(\displaystyle ky=2x-7\) is normal to the curve \(\displaystyle y=f(x)\) at \(\displaystyle (2,m)\).Find the value of \(\displaystyle k\) and \(\displaystyle m\).


any help will appreciate,.
There is not enough information given. Go back and check the question.
 
Oct 2009
202
4
There is not enough information given. Go back and check the question.

hi sir..

sorry,my mistake

the question is. A curve \(\displaystyle y=f(x)\) with gradient function \(\displaystyle \frac{dy}{dx}=\frac{x^2}{3x-1}\).
 

pickslides

MHF Helper
Sep 2008
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Well start by substituting \(\displaystyle x=2\) into \(\displaystyle \frac{dy}{dx}\) what you you get? What does that mean?
 
Oct 2009
202
4
Well start by substituting \(\displaystyle x=2\) into \(\displaystyle \frac{dy}{dx}\) what you you get? What does that mean?
hi,i get \(\displaystyle \frac{dy}{dx}=\frac{4}{5}\) then?it mean the gradient right?
 

pickslides

MHF Helper
Sep 2008
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hi,i get \(\displaystyle \frac{dy}{dx}=\frac{4}{5}\) then?it mean the gradient right?
Correct, it is the gradient of \(\displaystyle f(x)\) at \(\displaystyle x=2\). Now what is the relationship between the gradient and the normal?

Hint: \(\displaystyle m_N\times m_T = -1\)

How can we use this value?
 
Oct 2009
202
4
Correct, it is the gradient of \(\displaystyle f(x)\) at \(\displaystyle x=2\). Now what is the relationship between the gradient and the normal?

Hint: \(\displaystyle m_N\times m_T = -1\)

How can we use this value?
i get the gradient of normal is \(\displaystyle -\frac{5}{4}\).
 

pickslides

MHF Helper
Sep 2008
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Melbourne
The straight line \(\displaystyle ky=2x-7\) is normal to the curve \(\displaystyle y=f(x)\) at \(\displaystyle (2,m)\).
so maybe its' time to find \(\displaystyle k\)

Making the form \(\displaystyle y=mx+c\)

\(\displaystyle ky=2x-7\implies y=\frac{2}{k}x-\frac{7}{k}\)

\(\displaystyle \frac{-5}{4}= \frac{2}{k}\)
 
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