Find the tangent line to a curve
Find the tangent line to the curve $x^2y - y^2 + x = 11$ at the point $(3,1)$
I tried to solve it using parametric equations
begincases
y = t \[4px]
x = -dfrac12t + dfracsqrt1+4t^3 + 44t2t
endcases
and the derivative of $x(t)$, $y(t)$, $t= 1$ gives the direction vector $( 5/7 , 1 )$
so the line that passes through $(3,1)$
l: $(3,1) + s(5/7 , 1) $
Is it correct? Because using desmos the line doesn't seem to be tangent to the curve.
multivariable-calculus tangent-line
edited Nov 11 at 16:14
egreg
178k1484200
asked Nov 11 at 16:11
Razi Awad
226
add a comment |
Find the tangent line to the curve $x^2y - y^2 + x = 11$ at the point $(3,1)$
I tried to solve it using parametric equations
begincases
y = t \[4px]
x = -dfrac12t + dfracsqrt1+4t^3 + 44t2t
endcases
and the derivative of $x(t)$, $y(t)$, $t= 1$ gives the direction vector $( 5/7 , 1 )$
so the line that passes through $(3,1)$
l: $(3,1) + s(5/7 , 1) $
Is it correct? Because using desmos the line doesn't seem to be tangent to the curve.
multivariable-calculus tangent-line
edited Nov 11 at 16:14
egreg
178k1484200
asked Nov 11 at 16:11
Razi Awad
226
You have computed the derivative wrongly.
– egreg
Nov 11 at 16:29
That's not really using parametric equations to their full advantage. You've solved for x, and then used y=t to fake using parametric equations. You could also solve for y and then proceed as you normally would for y=f(x). But it would be better to approach this as an implicit or parametric equation, and use the methods there. That saves you from having to solve a quadratic, and from having to differentiate a fraction involving a square root.
– Teepeemm
Nov 11 at 20:42
add a comment |
Find the tangent line to the curve $x^2y - y^2 + x = 11$ at the point $(3,1)$
I tried to solve it using parametric equations
begincases
y = t \[4px]
x = -dfrac12t + dfracsqrt1+4t^3 + 44t2t
endcases
and the derivative of $x(t)$, $y(t)$, $t= 1$ gives the direction vector $( 5/7 , 1 )$
so the line that passes through $(3,1)$
l: $(3,1) + s(5/7 , 1) $
Is it correct? Because using desmos the line doesn't seem to be tangent to the curve.
multivariable-calculus tangent-line
edited Nov 11 at 16:14
egreg
178k1484200
asked Nov 11 at 16:11
Razi Awad
226
Find the tangent line to the curve $x^2y - y^2 + x = 11$ at the point $(3,1)$
I tried to solve it using parametric equations
begincases
y = t \[4px]
x = -dfrac12t + dfracsqrt1+4t^3 + 44t2t
endcases
and the derivative of $x(t)$, $y(t)$, $t= 1$ gives the direction vector $( 5/7 , 1 )$
so the line that passes through $(3,1)$
l: $(3,1) + s(5/7 , 1) $
Is it correct? Because using desmos the line doesn't seem to be tangent to the curve.
multivariable-calculus tangent-line
multivariable-calculus tangent-line
edited Nov 11 at 16:14
egreg
178k1484200
asked Nov 11 at 16:11
Razi Awad
226
edited Nov 11 at 16:14
egreg
178k1484200
asked Nov 11 at 16:11
Razi Awad
226
edited Nov 11 at 16:14
egreg
178k1484200
edited Nov 11 at 16:14
egreg
178k1484200
edited Nov 11 at 16:14
egreg
178k1484200
178k1484200
asked Nov 11 at 16:11
Razi Awad
226
asked Nov 11 at 16:11
Razi Awad
226
asked Nov 11 at 16:11
Razi Awad
226
226
You have computed the derivative wrongly.
– egreg
Nov 11 at 16:29
That's not really using parametric equations to their full advantage. You've solved for x, and then used y=t to fake using parametric equations. You could also solve for y and then proceed as you normally would for y=f(x). But it would be better to approach this as an implicit or parametric equation, and use the methods there. That saves you from having to solve a quadratic, and from having to differentiate a fraction involving a square root.
– Teepeemm
Nov 11 at 20:42
add a comment |
You have computed the derivative wrongly.
– egreg
Nov 11 at 16:29
That's not really using parametric equations to their full advantage. You've solved for x, and then used y=t to fake using parametric equations. You could also solve for y and then proceed as you normally would for y=f(x). But it would be better to approach this as an implicit or parametric equation, and use the methods there. That saves you from having to solve a quadratic, and from having to differentiate a fraction involving a square root.
– Teepeemm
Nov 11 at 20:42
You have computed the derivative wrongly.
– egreg
Nov 11 at 16:29
You have computed the derivative wrongly.
– egreg
Nov 11 at 16:29
That's not really using parametric equations to their full advantage. You've solved for x, and then used y=t to fake using parametric equations. You could also solve for y and then proceed as you normally would for y=f(x). But it would be better to approach this as an implicit or parametric equation, and use the methods there. That saves you from having to solve a quadratic, and from having to differentiate a fraction involving a square root.
– Teepeemm
Nov 11 at 20:42
That's not really using parametric equations to their full advantage. You've solved for x, and then used y=t to fake using parametric equations. You could also solve for y and then proceed as you normally would for y=f(x). But it would be better to approach this as an implicit or parametric equation, and use the methods there. That saves you from having to solve a quadratic, and from having to differentiate a fraction involving a square root.
– Teepeemm
Nov 11 at 20:42
add a comment |
2 Answers
2
active
oldest
votes
The tangent line at $(x_0,y_0)$ has equation
$$
(x-x_0)fracpartial fpartial x+
(y-y_0)fracpartial fpartial y=0
$$
where the partial derivatives are computed at $(x_0,y_0)$ and $f(x,y)=x^2y-y^2+x-11$.
Since
beginalign
fracpartial fpartial x&=2xy+1 \[6px]
fracpartial fpartial y&=x^2-2y
endalign
the tangent line has equation
$$
7(x-3)+7(y-1)=0
$$
that is, $x+y-4=0$.
You have made some error in computing the derivative of $x(t)$.
Solving the equation $x^2t+x-t^2-11=0$ with respect to $x$ yields
$$
x=frac-1pmsqrt1+44t+4t^32t
$$
If $t=1$, we get
$$
x=frac-1pm72
$$
so we need to take the branch with $+$.
The derivative of $x(t)$ is
$$
x'(t)=frac12cdotfracdfrac22+6t^2sqrt1+44t+4t^3-(sqrt1+44t+4t^3-1)t^2
$$
For $t=1$ we get
$$
x'(1)=frac12fracdfrac287-(7-1)1=-1
$$
answered Nov 11 at 16:19
egreg
178k1484200
Math and $mathrmLaTeX$... perfect combo!! $ddotsmile$.
– manooooh
Nov 11 at 16:26
thank you i have found the error but i love this way more its awesome glad to be half right
– Razi Awad
Nov 11 at 16:42
add a comment |
HINT
We have that
$$x^2y-y^2+x=11implies 2xydx+x^2dy-2ydy+dx=0impliesfracdydx=frac2xy+12y-x^2$$
then at $(3,1)$ the slope is $-1$.
answered Nov 11 at 16:21
gimusi
1
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The tangent line at $(x_0,y_0)$ has equation
$$
(x-x_0)fracpartial fpartial x+
(y-y_0)fracpartial fpartial y=0
$$
where the partial derivatives are computed at $(x_0,y_0)$ and $f(x,y)=x^2y-y^2+x-11$.
Since
beginalign
fracpartial fpartial x&=2xy+1 \[6px]
fracpartial fpartial y&=x^2-2y
endalign
the tangent line has equation
$$
7(x-3)+7(y-1)=0
$$
that is, $x+y-4=0$.
You have made some error in computing the derivative of $x(t)$.
Solving the equation $x^2t+x-t^2-11=0$ with respect to $x$ yields
$$
x=frac-1pmsqrt1+44t+4t^32t
$$
If $t=1$, we get
$$
x=frac-1pm72
$$
so we need to take the branch with $+$.
The derivative of $x(t)$ is
$$
x'(t)=frac12cdotfracdfrac22+6t^2sqrt1+44t+4t^3-(sqrt1+44t+4t^3-1)t^2
$$
For $t=1$ we get
$$
x'(1)=frac12fracdfrac287-(7-1)1=-1
$$
answered Nov 11 at 16:19
egreg
178k1484200
Math and $mathrmLaTeX$... perfect combo!! $ddotsmile$.
– manooooh
Nov 11 at 16:26
thank you i have found the error but i love this way more its awesome glad to be half right
– Razi Awad
Nov 11 at 16:42
add a comment |
The tangent line at $(x_0,y_0)$ has equation
$$
(x-x_0)fracpartial fpartial x+
(y-y_0)fracpartial fpartial y=0
$$
where the partial derivatives are computed at $(x_0,y_0)$ and $f(x,y)=x^2y-y^2+x-11$.
Since
beginalign
fracpartial fpartial x&=2xy+1 \[6px]
fracpartial fpartial y&=x^2-2y
endalign
the tangent line has equation
$$
7(x-3)+7(y-1)=0
$$
that is, $x+y-4=0$.
You have made some error in computing the derivative of $x(t)$.
Solving the equation $x^2t+x-t^2-11=0$ with respect to $x$ yields
$$
x=frac-1pmsqrt1+44t+4t^32t
$$
If $t=1$, we get
$$
x=frac-1pm72
$$
so we need to take the branch with $+$.
The derivative of $x(t)$ is
$$
x'(t)=frac12cdotfracdfrac22+6t^2sqrt1+44t+4t^3-(sqrt1+44t+4t^3-1)t^2
$$
For $t=1$ we get
$$
x'(1)=frac12fracdfrac287-(7-1)1=-1
$$
answered Nov 11 at 16:19
egreg
178k1484200
Math and $mathrmLaTeX$... perfect combo!! $ddotsmile$.
– manooooh
Nov 11 at 16:26
thank you i have found the error but i love this way more its awesome glad to be half right
– Razi Awad
Nov 11 at 16:42
add a comment |
The tangent line at $(x_0,y_0)$ has equation
$$
(x-x_0)fracpartial fpartial x+
(y-y_0)fracpartial fpartial y=0
$$
where the partial derivatives are computed at $(x_0,y_0)$ and $f(x,y)=x^2y-y^2+x-11$.
Since
beginalign
fracpartial fpartial x&=2xy+1 \[6px]
fracpartial fpartial y&=x^2-2y
endalign
the tangent line has equation
$$
7(x-3)+7(y-1)=0
$$
that is, $x+y-4=0$.
You have made some error in computing the derivative of $x(t)$.
Solving the equation $x^2t+x-t^2-11=0$ with respect to $x$ yields
$$
x=frac-1pmsqrt1+44t+4t^32t
$$
If $t=1$, we get
$$
x=frac-1pm72
$$
so we need to take the branch with $+$.
The derivative of $x(t)$ is
$$
x'(t)=frac12cdotfracdfrac22+6t^2sqrt1+44t+4t^3-(sqrt1+44t+4t^3-1)t^2
$$
For $t=1$ we get
$$
x'(1)=frac12fracdfrac287-(7-1)1=-1
$$
answered Nov 11 at 16:19
egreg
178k1484200
The tangent line at $(x_0,y_0)$ has equation
$$
(x-x_0)fracpartial fpartial x+
(y-y_0)fracpartial fpartial y=0
$$
where the partial derivatives are computed at $(x_0,y_0)$ and $f(x,y)=x^2y-y^2+x-11$.
Since
beginalign
fracpartial fpartial x&=2xy+1 \[6px]
fracpartial fpartial y&=x^2-2y
endalign
the tangent line has equation
$$
7(x-3)+7(y-1)=0
$$
that is, $x+y-4=0$.
You have made some error in computing the derivative of $x(t)$.
Solving the equation $x^2t+x-t^2-11=0$ with respect to $x$ yields
$$
x=frac-1pmsqrt1+44t+4t^32t
$$
If $t=1$, we get
$$
x=frac-1pm72
$$
so we need to take the branch with $+$.
The derivative of $x(t)$ is
$$
x'(t)=frac12cdotfracdfrac22+6t^2sqrt1+44t+4t^3-(sqrt1+44t+4t^3-1)t^2
$$
For $t=1$ we get
$$
x'(1)=frac12fracdfrac287-(7-1)1=-1
$$
answered Nov 11 at 16:19
egreg
178k1484200
edited Nov 11 at 16:33
answered Nov 11 at 16:19
egreg
178k1484200
answered Nov 11 at 16:19
egreg
178k1484200
answered Nov 11 at 16:19
egreg
178k1484200
178k1484200
Math and $mathrmLaTeX$... perfect combo!! $ddotsmile$.
– manooooh
Nov 11 at 16:26
thank you i have found the error but i love this way more its awesome glad to be half right
– Razi Awad
Nov 11 at 16:42
add a comment |
Math and $mathrmLaTeX$... perfect combo!! $ddotsmile$.
– manooooh
Nov 11 at 16:26
thank you i have found the error but i love this way more its awesome glad to be half right
– Razi Awad
Nov 11 at 16:42
Math and $mathrmLaTeX$... perfect combo!! $ddotsmile$.
– manooooh
Nov 11 at 16:26
Math and $mathrmLaTeX$... perfect combo!! $ddotsmile$.
– manooooh
Nov 11 at 16:26
thank you i have found the error but i love this way more its awesome glad to be half right
– Razi Awad
Nov 11 at 16:42
thank you i have found the error but i love this way more its awesome glad to be half right
– Razi Awad
Nov 11 at 16:42
add a comment |
HINT
We have that
$$x^2y-y^2+x=11implies 2xydx+x^2dy-2ydy+dx=0impliesfracdydx=frac2xy+12y-x^2$$
then at $(3,1)$ the slope is $-1$.
answered Nov 11 at 16:21
gimusi
1
add a comment |
HINT
We have that
$$x^2y-y^2+x=11implies 2xydx+x^2dy-2ydy+dx=0impliesfracdydx=frac2xy+12y-x^2$$
then at $(3,1)$ the slope is $-1$.
answered Nov 11 at 16:21
gimusi
1
add a comment |
HINT
We have that
$$x^2y-y^2+x=11implies 2xydx+x^2dy-2ydy+dx=0impliesfracdydx=frac2xy+12y-x^2$$
then at $(3,1)$ the slope is $-1$.
answered Nov 11 at 16:21
gimusi
1
HINT
We have that
$$x^2y-y^2+x=11implies 2xydx+x^2dy-2ydy+dx=0impliesfracdydx=frac2xy+12y-x^2$$
then at $(3,1)$ the slope is $-1$.
answered Nov 11 at 16:21
gimusi
1
answered Nov 11 at 16:21
gimusi
1
answered Nov 11 at 16:21
gimusi
1
answered Nov 11 at 16:21
gimusi
1
1
add a comment |
add a comment |
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You have computed the derivative wrongly.
– egreg
Nov 11 at 16:29
That's not really using parametric equations to their full advantage. You've solved for x, and then used y=t to fake using parametric equations. You could also solve for y and then proceed as you normally would for y=f(x). But it would be better to approach this as an implicit or parametric equation, and use the methods there. That saves you from having to solve a quadratic, and from having to differentiate a fraction involving a square root.
– Teepeemm
Nov 11 at 20:42