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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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






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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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            1300-talet

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            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output
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            Has there ever been an instance of an active nuclear power plant within or near a war zone?

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            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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