Find the equation of the Tangent Line to the given set of Parametric Equations at given point.
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I'm looking for validation for my answer to this question.
Parametric Equations: $x = t^2 + 2t + 1 , y = t^3 + 7t^2 + 8t, t = -1$
For this problem I used the Point-Slope-Form formula.
myAnswer:$ y = 19/2X + 17$
calculus parametric tangent-line
edited Nov 23 at 19:31
Key Flex
7,44941232
asked Nov 23 at 19:25
Alex Brito
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add a comment |
up vote
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down vote
favorite
I'm looking for validation for my answer to this question.
Parametric Equations: $x = t^2 + 2t + 1 , y = t^3 + 7t^2 + 8t, t = -1$
For this problem I used the Point-Slope-Form formula.
myAnswer:$ y = 19/2X + 17$
calculus parametric tangent-line
edited Nov 23 at 19:31
Key Flex
7,44941232
asked Nov 23 at 19:25
Alex Brito
14
At $t=-1$ we have $x=0$ and $y=-2$. Does this point lie on your line?
– amd
Nov 23 at 19:32
Where do I want to plug that t value
– Alex Brito
Nov 23 at 19:35
Would be better to detail your work.
– Yves Daoust
Nov 23 at 20:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm looking for validation for my answer to this question.
Parametric Equations: $x = t^2 + 2t + 1 , y = t^3 + 7t^2 + 8t, t = -1$
For this problem I used the Point-Slope-Form formula.
myAnswer:$ y = 19/2X + 17$
calculus parametric tangent-line
edited Nov 23 at 19:31
Key Flex
7,44941232
asked Nov 23 at 19:25
Alex Brito
14
I'm looking for validation for my answer to this question.
Parametric Equations: $x = t^2 + 2t + 1 , y = t^3 + 7t^2 + 8t, t = -1$
For this problem I used the Point-Slope-Form formula.
myAnswer:$ y = 19/2X + 17$
calculus parametric tangent-line
calculus parametric tangent-line
edited Nov 23 at 19:31
Key Flex
7,44941232
asked Nov 23 at 19:25
Alex Brito
14
edited Nov 23 at 19:31
Key Flex
7,44941232
asked Nov 23 at 19:25
Alex Brito
14
edited Nov 23 at 19:31
Key Flex
7,44941232
edited Nov 23 at 19:31
Key Flex
7,44941232
edited Nov 23 at 19:31
Key Flex
7,44941232
7,44941232
asked Nov 23 at 19:25
Alex Brito
14
asked Nov 23 at 19:25
Alex Brito
14
asked Nov 23 at 19:25
Alex Brito
14
14
At $t=-1$ we have $x=0$ and $y=-2$. Does this point lie on your line?
– amd
Nov 23 at 19:32
Where do I want to plug that t value
– Alex Brito
Nov 23 at 19:35
Would be better to detail your work.
– Yves Daoust
Nov 23 at 20:55
add a comment |
At $t=-1$ we have $x=0$ and $y=-2$. Does this point lie on your line?
– amd
Nov 23 at 19:32
Where do I want to plug that t value
– Alex Brito
Nov 23 at 19:35
Would be better to detail your work.
– Yves Daoust
Nov 23 at 20:55
At $t=-1$ we have $x=0$ and $y=-2$. Does this point lie on your line?
– amd
Nov 23 at 19:32
At $t=-1$ we have $x=0$ and $y=-2$. Does this point lie on your line?
– amd
Nov 23 at 19:32
Where do I want to plug that t value
– Alex Brito
Nov 23 at 19:35
Where do I want to plug that t value
– Alex Brito
Nov 23 at 19:35
Would be better to detail your work.
– Yves Daoust
Nov 23 at 20:55
Would be better to detail your work.
– Yves Daoust
Nov 23 at 20:55
add a comment |
2 Answers
2
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HINT
at $ t=-1 , (x_1,y_1)= (0,-2)$
$$ m=dfrac{dy}{dx}= dfrac{dy/dt}{dx/dt} $$
what is Point-Slope formula?
EDIT1:
slope calculation ( error corrected)
$$ dfrac{y+2}{x}=dfrac{3 t^2+14 t +8}{2t+2}=dfrac{3}{0}$$
answered Nov 23 at 19:39
Narasimham
20.6k52158
Point Slope Formula = (y -y1) = m (x-x1)
– Alex Brito
Nov 23 at 19:42
Can you apply it now?
– Narasimham
Nov 23 at 19:45
@AlexBrito You’re going to have a bit of trouble using the point-slope formula to find an equation for a vertical line.
– amd
Nov 23 at 19:57
@amd Are you referring to the parametric equations provided above?
– Alex Brito
Nov 23 at 19:58
@Narasimham I applied t = -1 in dx/dt and dy/dt. I got (0,-3)
– Alex Brito
Nov 23 at 19:59
|
show 2 more comments
up vote
0
down vote
The equation of the tangent at $t$ is
$$frac{x-f(t)}{f'(t)}=frac{y-g(t)}{g'(t)}.$$
With the given,
$$frac{x-0}{0}=frac{y+2}{3}.$$
As the left denominator is zero, the tangent must be vertical, and its equation reduces to
$$x=0.$$
answered Nov 23 at 21:00
Yves Daoust
123k668219
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
HINT
at $ t=-1 , (x_1,y_1)= (0,-2)$
$$ m=dfrac{dy}{dx}= dfrac{dy/dt}{dx/dt} $$
what is Point-Slope formula?
EDIT1:
slope calculation ( error corrected)
$$ dfrac{y+2}{x}=dfrac{3 t^2+14 t +8}{2t+2}=dfrac{3}{0}$$
answered Nov 23 at 19:39
Narasimham
20.6k52158
Point Slope Formula = (y -y1) = m (x-x1)
– Alex Brito
Nov 23 at 19:42
Can you apply it now?
– Narasimham
Nov 23 at 19:45
@AlexBrito You’re going to have a bit of trouble using the point-slope formula to find an equation for a vertical line.
– amd
Nov 23 at 19:57
@amd Are you referring to the parametric equations provided above?
– Alex Brito
Nov 23 at 19:58
@Narasimham I applied t = -1 in dx/dt and dy/dt. I got (0,-3)
– Alex Brito
Nov 23 at 19:59
|
show 2 more comments
up vote
1
down vote
HINT
at $ t=-1 , (x_1,y_1)= (0,-2)$
$$ m=dfrac{dy}{dx}= dfrac{dy/dt}{dx/dt} $$
what is Point-Slope formula?
EDIT1:
slope calculation ( error corrected)
$$ dfrac{y+2}{x}=dfrac{3 t^2+14 t +8}{2t+2}=dfrac{3}{0}$$
answered Nov 23 at 19:39
Narasimham
20.6k52158
Point Slope Formula = (y -y1) = m (x-x1)
– Alex Brito
Nov 23 at 19:42
Can you apply it now?
– Narasimham
Nov 23 at 19:45
@AlexBrito You’re going to have a bit of trouble using the point-slope formula to find an equation for a vertical line.
– amd
Nov 23 at 19:57
@amd Are you referring to the parametric equations provided above?
– Alex Brito
Nov 23 at 19:58
@Narasimham I applied t = -1 in dx/dt and dy/dt. I got (0,-3)
– Alex Brito
Nov 23 at 19:59
|
show 2 more comments
up vote
1
down vote
up vote
1
down vote
HINT
at $ t=-1 , (x_1,y_1)= (0,-2)$
$$ m=dfrac{dy}{dx}= dfrac{dy/dt}{dx/dt} $$
what is Point-Slope formula?
EDIT1:
slope calculation ( error corrected)
$$ dfrac{y+2}{x}=dfrac{3 t^2+14 t +8}{2t+2}=dfrac{3}{0}$$
answered Nov 23 at 19:39
Narasimham
20.6k52158
HINT
at $ t=-1 , (x_1,y_1)= (0,-2)$
$$ m=dfrac{dy}{dx}= dfrac{dy/dt}{dx/dt} $$
what is Point-Slope formula?
EDIT1:
slope calculation ( error corrected)
$$ dfrac{y+2}{x}=dfrac{3 t^2+14 t +8}{2t+2}=dfrac{3}{0}$$
answered Nov 23 at 19:39
Narasimham
20.6k52158
edited Nov 23 at 21:12
answered Nov 23 at 19:39
Narasimham
20.6k52158
answered Nov 23 at 19:39
Narasimham
20.6k52158
answered Nov 23 at 19:39
Narasimham
20.6k52158
20.6k52158
Point Slope Formula = (y -y1) = m (x-x1)
– Alex Brito
Nov 23 at 19:42
Can you apply it now?
– Narasimham
Nov 23 at 19:45
@AlexBrito You’re going to have a bit of trouble using the point-slope formula to find an equation for a vertical line.
– amd
Nov 23 at 19:57
@amd Are you referring to the parametric equations provided above?
– Alex Brito
Nov 23 at 19:58
@Narasimham I applied t = -1 in dx/dt and dy/dt. I got (0,-3)
– Alex Brito
Nov 23 at 19:59
|
show 2 more comments
Point Slope Formula = (y -y1) = m (x-x1)
– Alex Brito
Nov 23 at 19:42
Can you apply it now?
– Narasimham
Nov 23 at 19:45
@AlexBrito You’re going to have a bit of trouble using the point-slope formula to find an equation for a vertical line.
– amd
Nov 23 at 19:57
@amd Are you referring to the parametric equations provided above?
– Alex Brito
Nov 23 at 19:58
@Narasimham I applied t = -1 in dx/dt and dy/dt. I got (0,-3)
– Alex Brito
Nov 23 at 19:59
Point Slope Formula = (y -y1) = m (x-x1)
– Alex Brito
Nov 23 at 19:42
Point Slope Formula = (y -y1) = m (x-x1)
– Alex Brito
Nov 23 at 19:42
Can you apply it now?
– Narasimham
Nov 23 at 19:45
Can you apply it now?
– Narasimham
Nov 23 at 19:45
@AlexBrito You’re going to have a bit of trouble using the point-slope formula to find an equation for a vertical line.
– amd
Nov 23 at 19:57
@AlexBrito You’re going to have a bit of trouble using the point-slope formula to find an equation for a vertical line.
– amd
Nov 23 at 19:57
@amd Are you referring to the parametric equations provided above?
– Alex Brito
Nov 23 at 19:58
@amd Are you referring to the parametric equations provided above?
– Alex Brito
Nov 23 at 19:58
@Narasimham I applied t = -1 in dx/dt and dy/dt. I got (0,-3)
– Alex Brito
Nov 23 at 19:59
@Narasimham I applied t = -1 in dx/dt and dy/dt. I got (0,-3)
– Alex Brito
Nov 23 at 19:59
|
show 2 more comments
up vote
0
down vote
The equation of the tangent at $t$ is
$$frac{x-f(t)}{f'(t)}=frac{y-g(t)}{g'(t)}.$$
With the given,
$$frac{x-0}{0}=frac{y+2}{3}.$$
As the left denominator is zero, the tangent must be vertical, and its equation reduces to
$$x=0.$$
answered Nov 23 at 21:00
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
The equation of the tangent at $t$ is
$$frac{x-f(t)}{f'(t)}=frac{y-g(t)}{g'(t)}.$$
With the given,
$$frac{x-0}{0}=frac{y+2}{3}.$$
As the left denominator is zero, the tangent must be vertical, and its equation reduces to
$$x=0.$$
answered Nov 23 at 21:00
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
up vote
0
down vote
The equation of the tangent at $t$ is
$$frac{x-f(t)}{f'(t)}=frac{y-g(t)}{g'(t)}.$$
With the given,
$$frac{x-0}{0}=frac{y+2}{3}.$$
As the left denominator is zero, the tangent must be vertical, and its equation reduces to
$$x=0.$$
answered Nov 23 at 21:00
Yves Daoust
123k668219
The equation of the tangent at $t$ is
$$frac{x-f(t)}{f'(t)}=frac{y-g(t)}{g'(t)}.$$
With the given,
$$frac{x-0}{0}=frac{y+2}{3}.$$
As the left denominator is zero, the tangent must be vertical, and its equation reduces to
$$x=0.$$
answered Nov 23 at 21:00
Yves Daoust
123k668219
answered Nov 23 at 21:00
Yves Daoust
123k668219
answered Nov 23 at 21:00
Yves Daoust
123k668219
answered Nov 23 at 21:00
Yves Daoust
123k668219
123k668219
add a comment |
add a comment |
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At $t=-1$ we have $x=0$ and $y=-2$. Does this point lie on your line?
– amd
Nov 23 at 19:32
Where do I want to plug that t value
– Alex Brito
Nov 23 at 19:35
Would be better to detail your work.
– Yves Daoust
Nov 23 at 20:55