Tangent line to parametric curve at a point












1














Consider the curve given parametrically by $(x,y,z)=(2−t,−1−t^2,−2t−3t^3)$. There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(2,8,−162)$. Find the coordinates of the point $P$.



Can anyone give me a hand with this one? I honestly have no idea where to start other than rewriting the curve as,



$vec{r(t)}$ = $langle 2-t, -1-t^2, -2t-3t^3rangle$, and then taking the derivate to obtain,



$vec{r'(t)}$ = $langle -1, -2t, -2 -9t^2rangle$






parametric tangent-line







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  • What is an equation of the tangent line at $r(t)$? What does it mean in terms of this equation for $(2,8,-162)$ to be on this line?
    – amd
    May 19 '17 at 1:22










  • Managed to figure it out getting P equal to (5,-10,87). Thanks.
    – John
    May 19 '17 at 1:31
















1














Consider the curve given parametrically by $(x,y,z)=(2−t,−1−t^2,−2t−3t^3)$. There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(2,8,−162)$. Find the coordinates of the point $P$.



Can anyone give me a hand with this one? I honestly have no idea where to start other than rewriting the curve as,



$vec{r(t)}$ = $langle 2-t, -1-t^2, -2t-3t^3rangle$, and then taking the derivate to obtain,



$vec{r'(t)}$ = $langle -1, -2t, -2 -9t^2rangle$






parametric tangent-line







share|cite|improve this question






















  • What is an equation of the tangent line at $r(t)$? What does it mean in terms of this equation for $(2,8,-162)$ to be on this line?
    – amd
    May 19 '17 at 1:22










  • Managed to figure it out getting P equal to (5,-10,87). Thanks.
    – John
    May 19 '17 at 1:31














1












1








1







Consider the curve given parametrically by $(x,y,z)=(2−t,−1−t^2,−2t−3t^3)$. There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(2,8,−162)$. Find the coordinates of the point $P$.



Can anyone give me a hand with this one? I honestly have no idea where to start other than rewriting the curve as,



$vec{r(t)}$ = $langle 2-t, -1-t^2, -2t-3t^3rangle$, and then taking the derivate to obtain,



$vec{r'(t)}$ = $langle -1, -2t, -2 -9t^2rangle$






parametric tangent-line







share|cite|improve this question













Consider the curve given parametrically by $(x,y,z)=(2−t,−1−t^2,−2t−3t^3)$. There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(2,8,−162)$. Find the coordinates of the point $P$.



Can anyone give me a hand with this one? I honestly have no idea where to start other than rewriting the curve as,



$vec{r(t)}$ = $langle 2-t, -1-t^2, -2t-3t^3rangle$, and then taking the derivate to obtain,



$vec{r'(t)}$ = $langle -1, -2t, -2 -9t^2rangle$






parametric tangent-line




parametric tangent-line






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share|cite|improve this question











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asked May 19 '17 at 1:07









JohnJohn

549




549












  • What is an equation of the tangent line at $r(t)$? What does it mean in terms of this equation for $(2,8,-162)$ to be on this line?
    – amd
    May 19 '17 at 1:22










  • Managed to figure it out getting P equal to (5,-10,87). Thanks.
    – John
    May 19 '17 at 1:31


















  • What is an equation of the tangent line at $r(t)$? What does it mean in terms of this equation for $(2,8,-162)$ to be on this line?
    – amd
    May 19 '17 at 1:22










  • Managed to figure it out getting P equal to (5,-10,87). Thanks.
    – John
    May 19 '17 at 1:31
















What is an equation of the tangent line at $r(t)$? What does it mean in terms of this equation for $(2,8,-162)$ to be on this line?
– amd
May 19 '17 at 1:22




What is an equation of the tangent line at $r(t)$? What does it mean in terms of this equation for $(2,8,-162)$ to be on this line?
– amd
May 19 '17 at 1:22












Managed to figure it out getting P equal to (5,-10,87). Thanks.
– John
May 19 '17 at 1:31




Managed to figure it out getting P equal to (5,-10,87). Thanks.
– John
May 19 '17 at 1:31










2 Answers
2






active

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0














Once $vec{r'}=(-1,-2t,-2-9t^2)$ the tangent line is given by $$vec t=lambdavec{r'}text{, }lambda in Bbb R$$



So, we can write



$$vec P+vec t=(2,8,-162)to\
(2-t,-1-t^2,-2t-3t^3)+(-lambda,-2lambda t,-2lambda-9lambda t^2)=(2,8,-162)$$



which means:



$$2-t-lambda=2to t=-lambda$$



$$-1-t^2-2lambda t=8$$



$$-2t-3t^3-9lambda t^2=-162$$



Can you finish?






share|cite|improve this answer





















  • @John: Is it clear?
    – Arnaldo
    May 22 '17 at 17:27



















0














Let $vec{r}(t) = langle 2−t,−1−t^2,−2t−3t^3rangle$, where $tin mathbb{R}$. Then $vec{r}'(t) = langle -1, -2t, -2 -9t^2rangle$.



For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by
$$
vec{T}(t) = t :vec{r}'(t_0) + vec{r}(t_0), mbox{ where } tin mathbb{R}.
$$
Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$.



Thus, setting the following equal to each other,
begin{align*}
tlangle -1, -2t, -2 -9t^2rangle + langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3rangle = langle 2,8, -162 rangle,
end{align*}
we need to solve for $t_0$.



The above vector equation gives us three equations:
begin{align*}
-t+2-t_0&=2, hspace{4mm} (dagger) \
-2t_0t-1-t_0^2&=8, hspace{4mm} (ddagger) \
-2t-9t_0^2t-2t_0-3t_0^3&=-162. hspace{4mm} (Omega)\
end{align*}
Use the first equation $(dagger)$ to solve for $t:$
$$
t=-t_0.
$$
Let's substitute this into the second equation $(ddagger)$:
$$
2t_0^2 -1-t_0^2 = 8,
$$
which gives us $t_0^2 = 9$.
So $t_0 = pm 3$ while $t=mp 3$.



If $t_0=3$, then $t=-3$. We substitute these into the third equation $(Omega)$ to obtain:
$$
6+9cdot 9cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) not= -162.
$$
We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(Omega)$:
$$
-6-9cdot 9cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162.
$$
So $P$ has position vector $vec{r}(-3)=langle 5,-10,87 rangle $, which is when its tangent line passes through the point $(2,8,−162)$.






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    0














    Once $vec{r'}=(-1,-2t,-2-9t^2)$ the tangent line is given by $$vec t=lambdavec{r'}text{, }lambda in Bbb R$$



    So, we can write



    $$vec P+vec t=(2,8,-162)to\
    (2-t,-1-t^2,-2t-3t^3)+(-lambda,-2lambda t,-2lambda-9lambda t^2)=(2,8,-162)$$



    which means:



    $$2-t-lambda=2to t=-lambda$$



    $$-1-t^2-2lambda t=8$$



    $$-2t-3t^3-9lambda t^2=-162$$



    Can you finish?






    share|cite|improve this answer





















    • @John: Is it clear?
      – Arnaldo
      May 22 '17 at 17:27
















    0














    Once $vec{r'}=(-1,-2t,-2-9t^2)$ the tangent line is given by $$vec t=lambdavec{r'}text{, }lambda in Bbb R$$



    So, we can write



    $$vec P+vec t=(2,8,-162)to\
    (2-t,-1-t^2,-2t-3t^3)+(-lambda,-2lambda t,-2lambda-9lambda t^2)=(2,8,-162)$$



    which means:



    $$2-t-lambda=2to t=-lambda$$



    $$-1-t^2-2lambda t=8$$



    $$-2t-3t^3-9lambda t^2=-162$$



    Can you finish?






    share|cite|improve this answer





















    • @John: Is it clear?
      – Arnaldo
      May 22 '17 at 17:27














    0












    0








    0






    Once $vec{r'}=(-1,-2t,-2-9t^2)$ the tangent line is given by $$vec t=lambdavec{r'}text{, }lambda in Bbb R$$



    So, we can write



    $$vec P+vec t=(2,8,-162)to\
    (2-t,-1-t^2,-2t-3t^3)+(-lambda,-2lambda t,-2lambda-9lambda t^2)=(2,8,-162)$$



    which means:



    $$2-t-lambda=2to t=-lambda$$



    $$-1-t^2-2lambda t=8$$



    $$-2t-3t^3-9lambda t^2=-162$$



    Can you finish?






    share|cite|improve this answer












    Once $vec{r'}=(-1,-2t,-2-9t^2)$ the tangent line is given by $$vec t=lambdavec{r'}text{, }lambda in Bbb R$$



    So, we can write



    $$vec P+vec t=(2,8,-162)to\
    (2-t,-1-t^2,-2t-3t^3)+(-lambda,-2lambda t,-2lambda-9lambda t^2)=(2,8,-162)$$



    which means:



    $$2-t-lambda=2to t=-lambda$$



    $$-1-t^2-2lambda t=8$$



    $$-2t-3t^3-9lambda t^2=-162$$



    Can you finish?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 19 '17 at 1:34









    ArnaldoArnaldo

    18.1k42246




    18.1k42246












    • @John: Is it clear?
      – Arnaldo
      May 22 '17 at 17:27


















    • @John: Is it clear?
      – Arnaldo
      May 22 '17 at 17:27
















    @John: Is it clear?
    – Arnaldo
    May 22 '17 at 17:27




    @John: Is it clear?
    – Arnaldo
    May 22 '17 at 17:27











    0














    Let $vec{r}(t) = langle 2−t,−1−t^2,−2t−3t^3rangle$, where $tin mathbb{R}$. Then $vec{r}'(t) = langle -1, -2t, -2 -9t^2rangle$.



    For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by
    $$
    vec{T}(t) = t :vec{r}'(t_0) + vec{r}(t_0), mbox{ where } tin mathbb{R}.
    $$
    Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$.



    Thus, setting the following equal to each other,
    begin{align*}
    tlangle -1, -2t, -2 -9t^2rangle + langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3rangle = langle 2,8, -162 rangle,
    end{align*}
    we need to solve for $t_0$.



    The above vector equation gives us three equations:
    begin{align*}
    -t+2-t_0&=2, hspace{4mm} (dagger) \
    -2t_0t-1-t_0^2&=8, hspace{4mm} (ddagger) \
    -2t-9t_0^2t-2t_0-3t_0^3&=-162. hspace{4mm} (Omega)\
    end{align*}
    Use the first equation $(dagger)$ to solve for $t:$
    $$
    t=-t_0.
    $$
    Let's substitute this into the second equation $(ddagger)$:
    $$
    2t_0^2 -1-t_0^2 = 8,
    $$
    which gives us $t_0^2 = 9$.
    So $t_0 = pm 3$ while $t=mp 3$.



    If $t_0=3$, then $t=-3$. We substitute these into the third equation $(Omega)$ to obtain:
    $$
    6+9cdot 9cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) not= -162.
    $$
    We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(Omega)$:
    $$
    -6-9cdot 9cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162.
    $$
    So $P$ has position vector $vec{r}(-3)=langle 5,-10,87 rangle $, which is when its tangent line passes through the point $(2,8,−162)$.






    share|cite|improve this answer


























      0














      Let $vec{r}(t) = langle 2−t,−1−t^2,−2t−3t^3rangle$, where $tin mathbb{R}$. Then $vec{r}'(t) = langle -1, -2t, -2 -9t^2rangle$.



      For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by
      $$
      vec{T}(t) = t :vec{r}'(t_0) + vec{r}(t_0), mbox{ where } tin mathbb{R}.
      $$
      Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$.



      Thus, setting the following equal to each other,
      begin{align*}
      tlangle -1, -2t, -2 -9t^2rangle + langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3rangle = langle 2,8, -162 rangle,
      end{align*}
      we need to solve for $t_0$.



      The above vector equation gives us three equations:
      begin{align*}
      -t+2-t_0&=2, hspace{4mm} (dagger) \
      -2t_0t-1-t_0^2&=8, hspace{4mm} (ddagger) \
      -2t-9t_0^2t-2t_0-3t_0^3&=-162. hspace{4mm} (Omega)\
      end{align*}
      Use the first equation $(dagger)$ to solve for $t:$
      $$
      t=-t_0.
      $$
      Let's substitute this into the second equation $(ddagger)$:
      $$
      2t_0^2 -1-t_0^2 = 8,
      $$
      which gives us $t_0^2 = 9$.
      So $t_0 = pm 3$ while $t=mp 3$.



      If $t_0=3$, then $t=-3$. We substitute these into the third equation $(Omega)$ to obtain:
      $$
      6+9cdot 9cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) not= -162.
      $$
      We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(Omega)$:
      $$
      -6-9cdot 9cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162.
      $$
      So $P$ has position vector $vec{r}(-3)=langle 5,-10,87 rangle $, which is when its tangent line passes through the point $(2,8,−162)$.






      share|cite|improve this answer
























        0












        0








        0






        Let $vec{r}(t) = langle 2−t,−1−t^2,−2t−3t^3rangle$, where $tin mathbb{R}$. Then $vec{r}'(t) = langle -1, -2t, -2 -9t^2rangle$.



        For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by
        $$
        vec{T}(t) = t :vec{r}'(t_0) + vec{r}(t_0), mbox{ where } tin mathbb{R}.
        $$
        Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$.



        Thus, setting the following equal to each other,
        begin{align*}
        tlangle -1, -2t, -2 -9t^2rangle + langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3rangle = langle 2,8, -162 rangle,
        end{align*}
        we need to solve for $t_0$.



        The above vector equation gives us three equations:
        begin{align*}
        -t+2-t_0&=2, hspace{4mm} (dagger) \
        -2t_0t-1-t_0^2&=8, hspace{4mm} (ddagger) \
        -2t-9t_0^2t-2t_0-3t_0^3&=-162. hspace{4mm} (Omega)\
        end{align*}
        Use the first equation $(dagger)$ to solve for $t:$
        $$
        t=-t_0.
        $$
        Let's substitute this into the second equation $(ddagger)$:
        $$
        2t_0^2 -1-t_0^2 = 8,
        $$
        which gives us $t_0^2 = 9$.
        So $t_0 = pm 3$ while $t=mp 3$.



        If $t_0=3$, then $t=-3$. We substitute these into the third equation $(Omega)$ to obtain:
        $$
        6+9cdot 9cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) not= -162.
        $$
        We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(Omega)$:
        $$
        -6-9cdot 9cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162.
        $$
        So $P$ has position vector $vec{r}(-3)=langle 5,-10,87 rangle $, which is when its tangent line passes through the point $(2,8,−162)$.






        share|cite|improve this answer












        Let $vec{r}(t) = langle 2−t,−1−t^2,−2t−3t^3rangle$, where $tin mathbb{R}$. Then $vec{r}'(t) = langle -1, -2t, -2 -9t^2rangle$.



        For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by
        $$
        vec{T}(t) = t :vec{r}'(t_0) + vec{r}(t_0), mbox{ where } tin mathbb{R}.
        $$
        Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$.



        Thus, setting the following equal to each other,
        begin{align*}
        tlangle -1, -2t, -2 -9t^2rangle + langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3rangle = langle 2,8, -162 rangle,
        end{align*}
        we need to solve for $t_0$.



        The above vector equation gives us three equations:
        begin{align*}
        -t+2-t_0&=2, hspace{4mm} (dagger) \
        -2t_0t-1-t_0^2&=8, hspace{4mm} (ddagger) \
        -2t-9t_0^2t-2t_0-3t_0^3&=-162. hspace{4mm} (Omega)\
        end{align*}
        Use the first equation $(dagger)$ to solve for $t:$
        $$
        t=-t_0.
        $$
        Let's substitute this into the second equation $(ddagger)$:
        $$
        2t_0^2 -1-t_0^2 = 8,
        $$
        which gives us $t_0^2 = 9$.
        So $t_0 = pm 3$ while $t=mp 3$.



        If $t_0=3$, then $t=-3$. We substitute these into the third equation $(Omega)$ to obtain:
        $$
        6+9cdot 9cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) not= -162.
        $$
        We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(Omega)$:
        $$
        -6-9cdot 9cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162.
        $$
        So $P$ has position vector $vec{r}(-3)=langle 5,-10,87 rangle $, which is when its tangent line passes through the point $(2,8,−162)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 19 '17 at 1:49









        Mee Seong ImMee Seong Im

        2,7701517




        2,7701517






























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