# Homework Help: Three different answers.

1. Aug 4, 2011

### flyingpig

1. The problem statement, all variables and given/known data

Find the parametric equations of the tangent to the curve defined

$$r = <0, 2 + t^2, 3 + 2t - t^3>$$ at the point $$(0,2,3)$$

3. The attempt at a solution

a) $$r'(0) = <0, 0, 2>$$

$$r = <0, 2 + t^2, 3 + 2t - t^3>$$

$$r' = <0, 2 + 2t, 2 - 3t^2>$$

$$t = 0$$

$$r'(0) = <0,0,2>$$

b) q(u) = <0,2,0>

r = <0, 2 + t^2, 3 + 2t - t^3>

y = 2 + t^2 = 2

z = 3 + 2t - t^3 = 3

t = 0

Let y = f(z)

y' = \frac{2t}{2 - 3t^2}

y'(0) = 0

y - 2 = 0(z - 3) = 2[/tex]

Hence it is y = 2 or q(s) = <0,2,0>

c) $$p(r) = <0,2,3> + r<0,2,0>$$

$$r'(0) = <0, 0, 2>$$

r = <0, 2 + t^2, 3 + 2t - t^3>

r' = <0, 2 + 2t, 2 - 3t^2>

t = 0

r'(0) = <0,0,2>

p(r) = <0,2,3> + r<0,2,0>

I think a) is wrong because it is a point, not a point.

I can't decide between b) and c) because they both look correct with their reasoning. I lean towards c) though

2. Aug 4, 2011

### HallsofIvy

Was there a reason for putting your work into "spoilers"? It makes it very hard to read.

Do you know what "parametric equations" means?

There is an arithmetic error in (c).

3. Aug 4, 2011

Seconded.

4. Aug 4, 2011

### flyingpig

I have no taste in design.

parametric equation means x = x(t), y = y(t), z = z(t)? Ijust wrote it in vectors? Isn't that okay...?

I uploaded the file. I am kinda new to LaTeX, so I hope that is readable

File size:
66.3 KB
Views:
66
5. Aug 5, 2011

6. Aug 5, 2011

### HallsofIvy

Since you don't have any LaTeX in this post, I can't say about your LaTeX.

However, parametric equations for the tangent line to <f(t), g(t), h(t)> at the point $(x_0, y_0, z_0)$ are
$$x= x_0+ \frac{df}{dt}(t_0)(x- x_0)$$
$$y= y_0+ \frac{dg}{dt}(t_0)(y- y_0)$$
$$z= z_0+ \frac{dh}{dt}(t_0)(z- z_0)$$

Where $t_0$ is such that $x_0= f(t_0$, $y_0= g(t_0)$ and $z= h(t_0)$.

Equivalently, since the original curve was written as a vector function rather than parametric equations:
$$\left<x_0+ \frac{df}{dt}(t_0)(x- x_0), y= y_0+ \frac{dg}{dt}(t_0)(y- y_0), z_0+ \frac{dh}{dt}(t_0)(z- z_0)\right>$$

7. Aug 5, 2011

### flyingpig

Yeah, I know that, but I couldn't figure out why the other answers are wrong. Especially #2

I meant the file I uploaded. Is it not showing?

8. Aug 5, 2011

### SammyS

Staff Emeritus
Here's the Original Post with the spoilers revealed.
Where does q(u) come from? ... & many other questions.

9. Aug 5, 2011

### vela

Staff Emeritus
It's not so much that the file isn't showing. It's that what you've written is a bit incoherent. You say you know what a parametric equation is, yet solutions A and B suggest you don't. Where in those attempts is the answer the original question? There's nothing that looks like a parametric equation.

10. Aug 5, 2011

### Harrisonized

No idea what's going on with this one^
Spoilers are terrible here. They don't even look like real spoilers, where there's a button that upon clicking, reveals text inside a nice box. :(

Since you look absolutely clueless, the tangent curve can be found via:

r(t) = (t-t0)*(x'(t0),y'(t0),z'(t0)) + (x(t0),y(t0),z(t0))

where (x0,y0,z0) is what you equate to (x(t),y(t),z(t)) to obtain t0 from (x(t0),y(t0),z(t0)) where you want to find the tangent.

The first part (underlined) is the direction of the tangent vector. To make it extend to the entire line, you multiply it by t after plugging in the points. The second part (bolded) is necessary to shift the direction vector to the point on the curve that the parametric function runs through.

In your case, (x0,y0,z0) = (0,2,3). Therefore,

0=0 (for the x direction)
2=2+t0 (for the y direction)
3=3+2t0−t03 (for the z direction)

For the z direction, you have a special case:
0=2t0−t03
0=t0(2-t02)
t0=-sqrt(2),0,sqrt(2)

However, because of the parametric nature of this function, t0 must equal the same number in all three dimensions or the point doesn't exist on the function. Therefore, t0 must equal 0. Now all that's left is to -correctly- compute derivatives and plug in t0 to find your solution.

Last edited: Aug 5, 2011
11. Aug 6, 2011

### flyingpig

$$x = 0$$
$$y = 2$$
$$z = 3 + 2r$$

12. Aug 6, 2011

### vela

Staff Emeritus
It would really, really help in this thread and in others if you'd directly answer the questions asked of you instead of going off on another tangent. For example, I essentially asked you to explain what you're doing in your first two attempts. Instead of elaborating on what you did, you give the non-response of asking if the third attempt is correct.

So let's try again. Here's your solution A. Where is the parametric equation for the tangent?
What you wrote so far looks fine, but where's the rest?

Same thing for your solution B.
What's q? What's u? What's s? These things just seem to show up out of nowhere. How did you get your expression for y'? Again, where's the parametric equation that's supposedly your answer?

13. Aug 6, 2011

### flyingpig

q(u) is my new line with a new parameter u, I thought I make a new one so it won't confuse everyone...it did the opposite because I switched s and u.

The rest of the solution A is supposed to be in solution C.

My parametric equation is post 11

14. Aug 6, 2011

### vela

Staff Emeritus
OK, what you're saying is attempt A is only part of a solution and it's not independent of the others. So it's not an answer nor is it even a different answer, which is what you claimed in the title to this thread. Or were you saying r'(0)=(0,0,2) was your answer? Can you see why some of us were confused and wondered if you even knew what a parametric equation was?

Please explain what you're doing here:
Where did that come from? And why are you doing this?
Here, I think you just set t=0 in your previous expression.
Where did this come from? How can it possibly hold since 0 times anything is 0?
Again, how did you get y=2? How did you come up q?

15. Aug 7, 2011

### flyingpig

No they are supposed to be separate solutions. I just want to know why they are wrong.

basically I think it was doing this

r = <0, 2 + t^2, 3 - 2t - t^3>

y = 2 + t^2, y' = 2t
z = 3 - 2t - t^3, z' = -2 -3t^2

Since x = 0, everything is supposed to only be in yz-plane (I think that's what the solution is doing) and they decided to let y = f(z) so that y' = f'(z) = y'(t)/z'(t) = dy/dz

And that's what you see there.

y - 2 = 0(z - 3) = 2 was misleading, because I missed a > sign. It was supposed to be an arrow

y - 2 = 0(z - 3)
y - 2 = 0

y = 2

So "parametrically" it should be (a new parametric equation I am defining with a new parameter m) g(m) = <0,2,0>

16. Aug 7, 2011

### vela

Staff Emeritus
You honestly don't see a contradiction between those two statements, especially considering your "solution A" is repeated in "solution C"?

17. Aug 7, 2011

### vela

Staff Emeritus
OK, first, you should not use the same notation to denote different operations. If you use the prime to mean differentiation with respect to t, you need to write differentiation with respect to z in a different way. If you had simply written

dy/dz = y'(t)/z'(t) = 2t/(-2-3t2)

it would have made much more sense.

Second, simply stating the bolded phrase right from the start would have provided context to make sense of what you were doing. You should have said it in your original post.
So it was supposed to say y-2 = 0(z-3) => 2? That still wouldn't make sense because what is a lone "2" supposed to mean? You could have just left off that last piece.

Now, do you understand where the equation y - 2 = 0(z - 3) comes from?
This answer and the answer you gave in solution A suggest, again, you don't understand what a parametric equation of a line is. You should be able to write your answer in the form $$\vec{x}(t) = \vec{x}_0 + t\vec{d}$$
There are three pieces to this. First, t is the parameter. Second, $\vec{x}_0$ is the point the line goes through when the parameter t is equal to 0. Third, there's a non-zero vector $\vec{d}$ which specifies the direction the line goes in.

Your answer in A, r'(0) = (0,0,2), makes absolutely no sense. There's no parameter for one thing.

Your answer in B, g(m) = (0,2,0), isn't correct, though I can see now how you came up with it. In the yz-plane, the line y=2 is indeed the tangent to the curve. You just need to figure out how to write the equation of that line in parametric form. You already know the tangent passes through x0=(0,2,3), so your answer is going to look like$$\vec{x}(t) = (0, 2, 3) + t\vec{d}$$ You need to figure out the vector d that describes the direction of the line. Note that in your attempt, you assumed z=0, which is incorrect. With the line y=2 in the yz-plane, what values can z take on?

18. Aug 14, 2011

### flyingpig

It should be

$$\vec{x}(t) = <0,2,3> + t<0,2,3>$$

Sorry for the late reply came down with a cold.

With y = 2, it takes on all values of z.

19. Aug 14, 2011

### vela

Staff Emeritus
Try again.

20. Aug 15, 2011

### flyingpig

No I meant

$$\vec{x}(t) = <0,2,3> + t<0,0,2>$$