Differentiation how to find the equation of normal perpendicular to the straight line
$begingroup$
A curve has the equation $y=(2x-3)^2$. Find the equation of the normal to the curve that is perpendicular to the straight line $4x-y-5=0$,
I have differentiated the curve’s equation and I got $4(2x-3)$ as the answer. For the gradient of the straight line, I found that to be $4$. As it is perpendicular, the gradient for the normal would be $-frac{1}{4}$. Is that right? Well, after this, what are the following steps to find the the equation of the normal?
ordinary-differential-equations
edited Dec 4 '18 at 6:23
AVK
2,0961517
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
$endgroup$
add a comment |
$begingroup$
A curve has the equation $y=(2x-3)^2$. Find the equation of the normal to the curve that is perpendicular to the straight line $4x-y-5=0$,
I have differentiated the curve’s equation and I got $4(2x-3)$ as the answer. For the gradient of the straight line, I found that to be $4$. As it is perpendicular, the gradient for the normal would be $-frac{1}{4}$. Is that right? Well, after this, what are the following steps to find the the equation of the normal?
ordinary-differential-equations
edited Dec 4 '18 at 6:23
AVK
2,0961517
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
$endgroup$
add a comment |
$begingroup$
A curve has the equation $y=(2x-3)^2$. Find the equation of the normal to the curve that is perpendicular to the straight line $4x-y-5=0$,
I have differentiated the curve’s equation and I got $4(2x-3)$ as the answer. For the gradient of the straight line, I found that to be $4$. As it is perpendicular, the gradient for the normal would be $-frac{1}{4}$. Is that right? Well, after this, what are the following steps to find the the equation of the normal?
ordinary-differential-equations
edited Dec 4 '18 at 6:23
AVK
2,0961517
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
$endgroup$
A curve has the equation $y=(2x-3)^2$. Find the equation of the normal to the curve that is perpendicular to the straight line $4x-y-5=0$,
I have differentiated the curve’s equation and I got $4(2x-3)$ as the answer. For the gradient of the straight line, I found that to be $4$. As it is perpendicular, the gradient for the normal would be $-frac{1}{4}$. Is that right? Well, after this, what are the following steps to find the the equation of the normal?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 4 '18 at 6:23
AVK
2,0961517
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
edited Dec 4 '18 at 6:23
AVK
2,0961517
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
edited Dec 4 '18 at 6:23
AVK
2,0961517
edited Dec 4 '18 at 6:23
AVK
2,0961517
edited Dec 4 '18 at 6:23
AVK
2,0961517
2,0961517
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
asked Dec 4 '18 at 2:09
newbie2407newbie2407
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
The derivative gives you the slope of the tangent which is perpendicular to the normal - So the tangent must be parallel to the external line - as you correctly determined the tangent must have a slope of 4.
The normal has slope $-frac14$
and passes through the point $(
a, (2a-3)^2
)$
where $a$ is the solution to $4(2a-3)=4$
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The derivative gives you the slope of the tangent which is perpendicular to the normal - So the tangent must be parallel to the external line - as you correctly determined the tangent must have a slope of 4.
The normal has slope $-frac14$
and passes through the point $(
a, (2a-3)^2
)$
where $a$ is the solution to $4(2a-3)=4$
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
$endgroup$
add a comment |
$begingroup$
The derivative gives you the slope of the tangent which is perpendicular to the normal - So the tangent must be parallel to the external line - as you correctly determined the tangent must have a slope of 4.
The normal has slope $-frac14$
and passes through the point $(
a, (2a-3)^2
)$
where $a$ is the solution to $4(2a-3)=4$
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
$endgroup$
add a comment |
$begingroup$
The derivative gives you the slope of the tangent which is perpendicular to the normal - So the tangent must be parallel to the external line - as you correctly determined the tangent must have a slope of 4.
The normal has slope $-frac14$
and passes through the point $(
a, (2a-3)^2
)$
where $a$ is the solution to $4(2a-3)=4$
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
$endgroup$
The derivative gives you the slope of the tangent which is perpendicular to the normal - So the tangent must be parallel to the external line - as you correctly determined the tangent must have a slope of 4.
The normal has slope $-frac14$
and passes through the point $(
a, (2a-3)^2
)$
where $a$ is the solution to $4(2a-3)=4$
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
answered Dec 4 '18 at 2:47
WW1WW1
7,2951712
7,2951712
add a comment |
add a comment |
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