Equation of the normal to a curve [closed]
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I am struggling to find the equation of the normal to the line:
$$y = frac{1}{x} - frac{3}{x^2} - frac{4}{x^3} + frac{7}{4}$$ at $(-2,1)$. Any ideas would be appreciated. I believe I need to differentiate, but what do I do after that.
differential-equations curves
asked Nov 17 at 15:38
user8469209
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closed as unclear what you're asking by Henno Brandsma, Rebellos, Gibbs, José Carlos Santos, Leucippus Nov 18 at 1:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
0
down vote
favorite
I am struggling to find the equation of the normal to the line:
$$y = frac{1}{x} - frac{3}{x^2} - frac{4}{x^3} + frac{7}{4}$$ at $(-2,1)$. Any ideas would be appreciated. I believe I need to differentiate, but what do I do after that.
differential-equations curves
asked Nov 17 at 15:38
user8469209
82
closed as unclear what you're asking by Henno Brandsma, Rebellos, Gibbs, José Carlos Santos, Leucippus Nov 18 at 1:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
So you've got a function with a graph, with $(-2,1)$ on it. So you want the normal to the tangent at that point? Why cannot you differentiate? You do know the fomula for functions of the form $x^n$, which also works for $n<0$?
– Henno Brandsma
Nov 17 at 15:47
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am struggling to find the equation of the normal to the line:
$$y = frac{1}{x} - frac{3}{x^2} - frac{4}{x^3} + frac{7}{4}$$ at $(-2,1)$. Any ideas would be appreciated. I believe I need to differentiate, but what do I do after that.
differential-equations curves
asked Nov 17 at 15:38
user8469209
82
I am struggling to find the equation of the normal to the line:
$$y = frac{1}{x} - frac{3}{x^2} - frac{4}{x^3} + frac{7}{4}$$ at $(-2,1)$. Any ideas would be appreciated. I believe I need to differentiate, but what do I do after that.
differential-equations curves
differential-equations curves
asked Nov 17 at 15:38
user8469209
82
asked Nov 17 at 15:38
user8469209
82
edited Nov 17 at 15:59
asked Nov 17 at 15:38
user8469209
82
asked Nov 17 at 15:38
user8469209
82
asked Nov 17 at 15:38
user8469209
82
82
closed as unclear what you're asking by Henno Brandsma, Rebellos, Gibbs, José Carlos Santos, Leucippus Nov 18 at 1:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Henno Brandsma, Rebellos, Gibbs, José Carlos Santos, Leucippus Nov 18 at 1:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
So you've got a function with a graph, with $(-2,1)$ on it. So you want the normal to the tangent at that point? Why cannot you differentiate? You do know the fomula for functions of the form $x^n$, which also works for $n<0$?
– Henno Brandsma
Nov 17 at 15:47
add a comment |
So you've got a function with a graph, with $(-2,1)$ on it. So you want the normal to the tangent at that point? Why cannot you differentiate? You do know the fomula for functions of the form $x^n$, which also works for $n<0$?
– Henno Brandsma
Nov 17 at 15:47
So you've got a function with a graph, with $(-2,1)$ on it. So you want the normal to the tangent at that point? Why cannot you differentiate? You do know the fomula for functions of the form $x^n$, which also works for $n<0$?
– Henno Brandsma
Nov 17 at 15:47
So you've got a function with a graph, with $(-2,1)$ on it. So you want the normal to the tangent at that point? Why cannot you differentiate? You do know the fomula for functions of the form $x^n$, which also works for $n<0$?
– Henno Brandsma
Nov 17 at 15:47
add a comment |
1 Answer
1
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1
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Tips:
- Remember $;biggl(dfrac1{x^n}biggr)'=-dfrac{n}{x^{n+1}},;$ so
$$y'=-frac2{x^2}+frac{6}{x^3}+frac{12}{x^4}. $$
- An equation of the straight line with given slope $m$, passing through a given point $(x_0,y_0)$ is given by the formula
$$y-y_0=m(x-x_0).$$
The normal at the point with abscissa $-2$ will have slope $m=dfrac{-1}{y'(-2)}$, and its equation will be
$$y-1=m(x+2).$$
answered Nov 17 at 15:59
Bernard
116k637108
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Tips:
- Remember $;biggl(dfrac1{x^n}biggr)'=-dfrac{n}{x^{n+1}},;$ so
$$y'=-frac2{x^2}+frac{6}{x^3}+frac{12}{x^4}. $$
- An equation of the straight line with given slope $m$, passing through a given point $(x_0,y_0)$ is given by the formula
$$y-y_0=m(x-x_0).$$
The normal at the point with abscissa $-2$ will have slope $m=dfrac{-1}{y'(-2)}$, and its equation will be
$$y-1=m(x+2).$$
answered Nov 17 at 15:59
Bernard
116k637108
add a comment |
up vote
1
down vote
Tips:
- Remember $;biggl(dfrac1{x^n}biggr)'=-dfrac{n}{x^{n+1}},;$ so
$$y'=-frac2{x^2}+frac{6}{x^3}+frac{12}{x^4}. $$
- An equation of the straight line with given slope $m$, passing through a given point $(x_0,y_0)$ is given by the formula
$$y-y_0=m(x-x_0).$$
The normal at the point with abscissa $-2$ will have slope $m=dfrac{-1}{y'(-2)}$, and its equation will be
$$y-1=m(x+2).$$
answered Nov 17 at 15:59
Bernard
116k637108
add a comment |
up vote
1
down vote
up vote
1
down vote
Tips:
- Remember $;biggl(dfrac1{x^n}biggr)'=-dfrac{n}{x^{n+1}},;$ so
$$y'=-frac2{x^2}+frac{6}{x^3}+frac{12}{x^4}. $$
- An equation of the straight line with given slope $m$, passing through a given point $(x_0,y_0)$ is given by the formula
$$y-y_0=m(x-x_0).$$
The normal at the point with abscissa $-2$ will have slope $m=dfrac{-1}{y'(-2)}$, and its equation will be
$$y-1=m(x+2).$$
answered Nov 17 at 15:59
Bernard
116k637108
Tips:
- Remember $;biggl(dfrac1{x^n}biggr)'=-dfrac{n}{x^{n+1}},;$ so
$$y'=-frac2{x^2}+frac{6}{x^3}+frac{12}{x^4}. $$
- An equation of the straight line with given slope $m$, passing through a given point $(x_0,y_0)$ is given by the formula
$$y-y_0=m(x-x_0).$$
The normal at the point with abscissa $-2$ will have slope $m=dfrac{-1}{y'(-2)}$, and its equation will be
$$y-1=m(x+2).$$
answered Nov 17 at 15:59
Bernard
116k637108
edited Nov 17 at 16:04
answered Nov 17 at 15:59
Bernard
116k637108
answered Nov 17 at 15:59
Bernard
116k637108
answered Nov 17 at 15:59
Bernard
116k637108
116k637108
add a comment |
add a comment |
So you've got a function with a graph, with $(-2,1)$ on it. So you want the normal to the tangent at that point? Why cannot you differentiate? You do know the fomula for functions of the form $x^n$, which also works for $n<0$?
– Henno Brandsma
Nov 17 at 15:47