Question

Problem:
A curve given parametrically by (x, y, z) = (2 + 3t, 2 – 2t^2, -3t – 2t^3). There is a unique point P on the curve with the property that the tangent line at P passes through the point (-10, -22, 76).

Answer:
P = (-4, -6, 22)

I would appreciate it if someone could please tell me how to arrive at that answer!

Answer 1

are you familiar with the formula for a parametric line\[\vec r(t)=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle\]?

Answer 2

I had forgotten about it, but you refreshed my mind, I suppose. :)

Answer 3

\(\langle a,b,c\rangle\) is a vector along the line, and \(\langle x_0,y_0,z_0\rangle\) is a point on the line. hopefully you can see how this traces out a line in space parametrically as t increases

Answer 4

for more detail:
http://tutorial.math.lamar.edu/Classes/CalcII/EqnsOfLines.aspx

Answer 5

in your problem, you have to find the initial point, given the line. from the equation of the line we can get our vector that points along the line tangentially by taking the derivative

Answer 6

\[\vec r'(t)=\langle 3,-4t,-3-6t^2\rangle\]which in this case is your \[\langle a,b,c\rangle\]

Answer 7

(-10, -22, 76) = (x_0, y_0, z_0)?

Answer 8

Actually, wait.

Answer 9

(x_0, y_0, z_0) is the y-intercept of the regular line, and (-10, -22, 76) is the y-intercept of the tangent line, right?

Answer 10

Or, rather, the "anchor".

Answer 11

Not intercept, right?

Answer 12

nor sure why you think this has to do with the y-intercept
your \(\langle x_0,y_0,z_0\rangle\) is on your given parametric line...

Answer 13

\(\langle x_0,y_0,z_0\rangle\) is actually what we're looking for
since we know it's on the line we can write\[\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle=\vec r(t)+t\vec r'(t)\]

Answer 14

since we know that this line goes through point P for some t, we then can write\[\vec P=\vec r(t)+t\vec r'(t)\]

Answer 15

Is <x_o, y_0, z_0> a point that's on r(t) and r'(t) or just r'(t)?

Answer 16

there is no such thing as a point "on" r'(t), r'(t) is a function that describes the way the tangent arrow on the curve r(t) faces at any given time. it cannot be said to "trace out a curve" like r(t) can.
so the answer is that <x_0, y_0, z_0> is a point that's on r(t)

Answer 17

Ok, that makes sense, but let me think about the other stuff you said. 2 secs.

Answer 18

sure thing :)

Answer 19

take your time, I'm multitasking :P

Answer 20

Alright, lol. :)

Answer 21

So <a,b,c> = <c_1, c_2, c_2> where c_n is each component of dr/dt?

Ok, so, I think I see that P = r + tr'. I'm going to do the work with my digital pen, and I'll get back to you. :)

Answer 22

Actually, wait.

Answer 23

Is r(t) = (x,y,z) or P + r(t) = (x,y,z)?

Answer 24

Wait, let me think more before I ask something, just in case it's a stupid question.

Answer 25

(I slept a little poorly today.)

Answer 26

we didn't talk about x,y,z yet, but P is the point we are "pointing to", so your first statement was accurate

Answer 27

Is this "Is r(t) = (x,y,z)" what you're saying is accurate?

Answer 28

such is college life :p
<a,b,c> = <c_1, c_2, c_2> where c_n is each component of dr/dt - is correct

Answer 29

Oh, and, sadly, it is. :(

Answer 30

r(t) is given, we want a specific point *on* r(t) that we called <x_0,y_0,z_0>

Answer 31

Okay, so, r(t) IS equal to (x,y,z)?

Answer 32

(Not saying (x_0,y_0,z_0).)

Answer 33

I don't know what you really mean, but I'm going to say "no"
let's say <x,y,z> is the line we are trying to find to be clear, call it something else, like v(t)
r(t) is the curve we are given
<x_0, y_0, z_0> is a point on that curve

Answer 34

if it helps you to think of it this way, the equation of the line we want is\[\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle=\vec r(t)+t\vec r'(t)=\langle x,y,z\rangle=\vec v(t)\]where v(t) is the line we need to find the equation of

Answer 35

on the line v(t) is the point P somewhere, so we can say that there exist some x,y,z such that for some value of t, we have\[\vec v(t)=\vec P=\langle x,y,z\rangle=\langle-10,-22,76\rangle=\vec r(t)+t\vec r'(t)\]

Answer 36

Actually, you lost me with that.

I can't relate that to r = P + tr'.

I'm really sorry. :(

Answer 37

solve for t

Answer 38

Wait, you seem to contradict yourself.

Answer 39

P = <x,y,z>?

I thought P was <a,b,c>.

<a,b,c> = <x,y,z>?

Answer 40

If P = <x,y,z> then r = P + tr' doesn't hold.

Answer 41

no, <a,b,c> is r'(t) for some t

Answer 42

it is the tangent vector we need to get our line

Answer 43

I meant P = <x_0, y_0, z_0>, I think.

Answer 44

Wait, I'm really sorry, I'm kind of a mess, and I'm trying really hard to focus! :@

Answer 45

Oh, wait!

Answer 46

that's not true either, and no worries about it, my explanation may not be the best..
in general, the equation of a line parametrically is\[\vec v(t)=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle\]right?

Answer 47

sorry my internet went out for a sec

Answer 48

had a revelation?

Answer 49

I did, but I lost it kind of. Basically, what do we do after we have t?

Answer 50

the question is asking for the point on the curve that has the tangent line we found, so we want to plug in the value of t into r(t)

Answer 51

Okay, so, I'm going to rewrite this all concisely, because of my sleep-deprived state when I scroll up and down too much, I forget what I was thinking.

Answer 52

haha okay, let me try to write out a concise explanation in the meantime...

Answer 53

:$

Answer 54

Is r'(t) technically the derivative of r(t) - (x_0, y_0, z_0) which just happens to be the derivative of r(t)?

Answer 55

Wait, I take that back for now.

Answer 56

ok lol

Answer 57

:D

Answer 58

The equation of a line parametrically is\[\vec v(t)=\langle x,y,z\rangle=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle\]where \(\langle\ x_0,y_0,z_0\rangle\) is a point on the line, and \(\langle\ a,b,c\rangle\) is a vector in the same direction as the line \(\vec v(t)\).
Note: do NOT confuse this with the equation of the given curve \(\vec r(t)\). I used the same variables at the beginning of my explanation and that probably did not help...
In this case, \(\langle\ x_0,y_0,z_0\rangle\) is on the curve given, \(\vec r(t)\), and the line must be tangent to the curve, and so must point in the direction of \(\langle a,b,c\rangle=\vec r'(t)\)
Putting all this together, this means we can write the equation of the tangent line to the curve we are looking for as\[\vec v(t)=\vec r(t)+t\vec r'(t)\]
Now all we need to do is note that \(P\) is on \(\vec v(t)\) and we can write\[\vec P=\vec r(t)+t\vec r'(t)\]which can be solved for \(t\)

Once you have that, you can get the point on the curve we started from, \(\langle x_0,y_0,z_0\rangle\), but plugging that value of \(t\) back into the equation of the curve \(\vec r(t)\), which gives you your answer.

Answer 59

Ok, so, (2 + 3t, 2 – 2t^2, -3t – 2t^3) = (-10, -22, 76) yields different values of t per component.

Answer 60

you can't set the equation for the curve equal to the point; the point is not on the curve
read what I wrote above and let me know where I lose you if you could :P

Answer 61

But, you said ⟨x,y,z⟩=⟨x0,y0,z0⟩+t⟨a,b,c⟩ = P

I took P = (-10, -22, 76) and equated it to P = (x,y,z) = v(t).

Answer 62

yeah that's right

Answer 63

(x,y,z) = (2 + 3t, 2 – 2t^2, -3t – 2t^3)

so, (-10, -22, 76) = (2 + 3t, 2 – 2t^2, -3t – 2t^3), but that gives me different values of t per component.

Answer 64

no, you are again confusing the equation of the line and that of the curve r(t)

Answer 65

<x,y,z>=v(t) NOT r(t)

Answer 66

r(t) = (2 + 3t, 2 – 2t^2, -3t – 2t^3) ?

if so, why does the problem say (x, y, z) = (2 + 3t, 2 – 2t^2, -3t – 2t^3)?

Answer 67

<x,y,z>=r(t_0) + r'(t_0) for some t = t_0 as per my argument above

Answer 68

Okay, I think I should just do re-read this tomorrow. :P

Answer 69

okay, terminology issue is all that is
fine let's call the function r(t)=<x,y,z> like they do (which I didn't realize, sorry)
then we need another name for the components in the line we are trying to find, so then we can call those v(t)=<p,q,r>
the semantics are what's killing us here :P

Answer 70

so let's say then that the line can be written v(t)=<p,q,r>, then we'll say
v(t)=r(t)+t*r'(t)

<p,q,r>=<x,y,z>+t<a,b,c>

and for some particular value t=t_0, we can call the point

<p_0,q_0,r_0>=P=<x_0,y_0,z_0>+t<a,b,c>

Answer 71

change the line in my above explanation and say the general equation of a line is\[\vec v(t)=\langle p,q,r\rangle=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle\]if that helps...

Answer 72

Actually, I, honestly, think I'm too mentally drained to complete this problem successfully today, but I have a feeling that the last few posts will have cleared things up. I'll likely check it tomorrow, and, hopefully, report that I get it. :)

Answer 73

Thanks a lot for your explanations.

Answer 74

sure thing, that often helps
good luck, get some shut-eye!

Answer 75

I intend to. :D

See you some other time. Bye for now!

Answer 76

@TuringTest:

I changed the variables to be better compatible with my brain. :) I get the right answer, but did I get it correctly, or is it a fluke?:

(x,y,z) = (2 + 3t, 2 - 2t^2, -3t - 2t^3)
d/dt (x,y,z) = (3, -4t, -3, -6t^2)

Let's call the point (-10,-22, 76) point Q.:

Q = (-10,-22,76)

Since we know that both the curve and one of its tangent lines pass through a point P and that that same tangent line also passes through the point we denoted by Q, the following must hold.:
Q = (x,y,z) + t * d/dt (x,y,z)
(-10,-22,76) = (2 + 3t, 2 - 2t^2, -3t - 2t^3) + t * (3, -4t, -3 - 6t^2)
(-10,-22,76) = (2 + 3t + 3t, 2 - 2t^2 - 4t^2, -3t - 3t - 2t^3 - 6t^3)
(-10,-22,76) = (2 + 6t, 2 - 6t^2, - 6t - 8t^3)

From this we find the following three equations.:
-10 = 2 + 6t ==> t = -12/6 = -2
-22 = 2 - 6t^2 ==> t = -2 holds for this equation too.
76 = -6t - 8t^3 ==> t = -2 holds for this equation too.

Now that we know that t = -2, we can plug it back into (x,y,z) = (2 + 3t, 2 - 2t^2, -3t - 2t^3) to get (2 + 3*(-2), 2 - 2*(-2)^2, -3*(-2) - 2*(-2)^3) = (-4,-6,22).

Answer 77

I should note that when plugging back in -2 to (x,y,z), I'm plugging in back t_P = -2. (Emphasis on the subscript P of t_P.)

Answer 78

yep, you got it :)
just needed a little rest I guess

Answer 79

Thanks, again. :)