Find the equation to the tangent of a line using known points?
up vote
1
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I have carried out the implicit differentiation of the original formula ($x-y^3=2xy$) to get the equation
$$frac{dy}{dx} = - frac{2y-1}{3y^2-2x}.$$
Now I need to find the equation of the tangent at point $(-1, 1)$, I've plugged the values into the formula to get
$$frac{dy}{dx} = - frac15$$
but have a suspicion I may be missing something.
Many thanks!
derivatives
edited Nov 14 at 16:19
Tianlalu
2,599632
asked Nov 14 at 15:59
RocketKangaroo
163
add a comment |
up vote
1
down vote
favorite
I have carried out the implicit differentiation of the original formula ($x-y^3=2xy$) to get the equation
$$frac{dy}{dx} = - frac{2y-1}{3y^2-2x}.$$
Now I need to find the equation of the tangent at point $(-1, 1)$, I've plugged the values into the formula to get
$$frac{dy}{dx} = - frac15$$
but have a suspicion I may be missing something.
Many thanks!
derivatives
edited Nov 14 at 16:19
Tianlalu
2,599632
asked Nov 14 at 15:59
RocketKangaroo
163
You have a sign error in the denominator.
– amd
Nov 14 at 20:18
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have carried out the implicit differentiation of the original formula ($x-y^3=2xy$) to get the equation
$$frac{dy}{dx} = - frac{2y-1}{3y^2-2x}.$$
Now I need to find the equation of the tangent at point $(-1, 1)$, I've plugged the values into the formula to get
$$frac{dy}{dx} = - frac15$$
but have a suspicion I may be missing something.
Many thanks!
derivatives
edited Nov 14 at 16:19
Tianlalu
2,599632
asked Nov 14 at 15:59
RocketKangaroo
163
I have carried out the implicit differentiation of the original formula ($x-y^3=2xy$) to get the equation
$$frac{dy}{dx} = - frac{2y-1}{3y^2-2x}.$$
Now I need to find the equation of the tangent at point $(-1, 1)$, I've plugged the values into the formula to get
$$frac{dy}{dx} = - frac15$$
but have a suspicion I may be missing something.
Many thanks!
derivatives
derivatives
edited Nov 14 at 16:19
Tianlalu
2,599632
asked Nov 14 at 15:59
RocketKangaroo
163
edited Nov 14 at 16:19
Tianlalu
2,599632
asked Nov 14 at 15:59
RocketKangaroo
163
edited Nov 14 at 16:19
Tianlalu
2,599632
edited Nov 14 at 16:19
Tianlalu
2,599632
edited Nov 14 at 16:19
Tianlalu
2,599632
2,599632
asked Nov 14 at 15:59
RocketKangaroo
163
asked Nov 14 at 15:59
RocketKangaroo
163
asked Nov 14 at 15:59
RocketKangaroo
163
163
You have a sign error in the denominator.
– amd
Nov 14 at 20:18
add a comment |
You have a sign error in the denominator.
– amd
Nov 14 at 20:18
You have a sign error in the denominator.
– amd
Nov 14 at 20:18
You have a sign error in the denominator.
– amd
Nov 14 at 20:18
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
You have the slope of the line and a point on the line.
Use the slope point form of line equation
$$y=mx+c,$$
where $m$ is the slope,
then plug in the point to get $c$.
edited Nov 14 at 16:20
Tianlalu
2,599632
answered Nov 14 at 16:04
user199996
1
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You have the slope of the line and a point on the line.
Use the slope point form of line equation
$$y=mx+c,$$
where $m$ is the slope,
then plug in the point to get $c$.
edited Nov 14 at 16:20
Tianlalu
2,599632
answered Nov 14 at 16:04
user199996
1
add a comment |
up vote
0
down vote
You have the slope of the line and a point on the line.
Use the slope point form of line equation
$$y=mx+c,$$
where $m$ is the slope,
then plug in the point to get $c$.
edited Nov 14 at 16:20
Tianlalu
2,599632
answered Nov 14 at 16:04
user199996
1
add a comment |
up vote
0
down vote
up vote
0
down vote
You have the slope of the line and a point on the line.
Use the slope point form of line equation
$$y=mx+c,$$
where $m$ is the slope,
then plug in the point to get $c$.
edited Nov 14 at 16:20
Tianlalu
2,599632
answered Nov 14 at 16:04
user199996
1
You have the slope of the line and a point on the line.
Use the slope point form of line equation
$$y=mx+c,$$
where $m$ is the slope,
then plug in the point to get $c$.
edited Nov 14 at 16:20
Tianlalu
2,599632
answered Nov 14 at 16:04
user199996
1
edited Nov 14 at 16:20
Tianlalu
2,599632
edited Nov 14 at 16:20
Tianlalu
2,599632
edited Nov 14 at 16:20
Tianlalu
2,599632
2,599632
answered Nov 14 at 16:04
user199996
1
answered Nov 14 at 16:04
user199996
1
answered Nov 14 at 16:04
user199996
1
1
add a comment |
add a comment |
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You have a sign error in the denominator.
– amd
Nov 14 at 20:18