# Homework Help: X intercept of Line Tangent to Curve

1. May 20, 2014

### mill

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

What is the x-intercept of the line tangent to the curve x(t) = 3 + cos(∏t), y(t) = t^2 + t + 1, when t = 1?

2. Relevant equations

Derivative, y=mx+b

3. The attempt at a solution

To find the line tangent to the curve:

d/dt = <-∏sin(∏t), 2t+1>

at t=1 <-∏, 3>

dy/dx = dy/du * du/dx = -3/∏

y=(-3/∏)x+b

at t=1, <x,y>=<3,3>

y-3 = (-3/∏)(x-3)

y = -3x /∏ + 9/∏ +3

To find x-int. set y=0 so 0=-3x /∏ + 9/∏ +3

(-3 - 9/∏)(-∏/3)=x

The answer is x=2. I am not sure where I went wrong.

Last edited: May 20, 2014
2. May 20, 2014

### LCKurtz

$\sin(\pi)=0$

What is $u$? I think you mean $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. But you could just use the parametric equation of the tangent line instead of using $y=mx+b$ form.

3. May 20, 2014

### mill

I see sin(∏)=0. Then that would mean dy/dx = 0.

By parametric equation of the tangent line do you mean r'(t)=x' + y'? To find int. I would set r'(1)=x' + y' and x'=-∏sin(t∏). t=1 so x'=-∏sin(∏)=0. y'=3

x'=sin(∏)=0 Inv sin (0)=0

r'(t)=3

4. May 20, 2014

### LCKurtz

No. The 0 is in the denominator, so the slope is undefined. All the more reason not to try using $y=mx+b$.

That is really poor notation. $r(t) = \langle x(t),y(t)\rangle$ which is a vector, not a sum, and $r'(t) = \langle x'(t),y'(t)\rangle$ so $r'(1) = \langle -\pi\sin\pi,3\rangle=\langle 0,3\rangle$. Since $r(1) = \langle 2,3\rangle$, you have a point and direction vector for the tangent line when $t=1$. What is the parametric equation of that tangent line?

5. May 20, 2014

### mill

Would it be r(t)=<0, 3> + t<2,3> for the tangent? I am not sure how I can say the intercept is two from this equation. If I just look at x then x=2t which at t=1 is 2. Is that how it is found?

6. May 20, 2014

### LCKurtz

Close but you have it backwards. <2,3> is the point and <0,3> is the direction vector. And it's best to use a different parameter and a different name so $L(s) = \langle<x(s),y(s)\rangle=\langle 2,3\rangle + s\langle 0,3\rangle$, where $L$ is the tangent line.

What value of $s$ makes $y(s)=0$? What is $x(s)$ for that value.

Also, try not to lose sight of the forest for the trees. In the xy plane you have a line through (2,3) and its direction vector is <0,3>. What direction is that? What does the line look like?

7. May 20, 2014

### mill

s would need to be 0. Would I necessarily need to find y in order to find the x-intercept?

The point (2,3) is going towards (0,3) so it is a line headed downwards along x?

8. May 20, 2014

### LCKurtz

Why?

The x intercept on any graph is the x value when y = 0.

I don't know what "(2,3) going towards (0,3)" means nor what "headed downwards along x" means. The graph is a straight line. Surely you can describe it better than that. Plot a few points for various s.

9. May 20, 2014

### mill

If I separate them into x=2+s(0) and y=3+s(3) then the only thing that makes y=0 is s=0?

10. May 20, 2014

### LCKurtz

Are you serious?

11. May 20, 2014

### mill

If I weren't, I wouldn't still be here eight hours later, sadly. Going off your equation L(s) =<x(s), y(s)> I assumed that I can simply look at x(s) since it is the x-intercept. s=0 so x=2. If I'm missing something, I don't know what it is.

12. May 20, 2014

### LCKurtz

You have y = 3+3s and you are saying y = 0 when s = 0???

In this problem x happens to be constant so x=2 is the correct intercept even though you have the incorrect value of s for y=0. Can you describe what the line looks like?