# Thread: Velocity and Normal line problems

1. ## Velocity and Normal line problems

1.Suppose that "f" is an objects position at time "x" in seconds.

x is greater than or equal to 0 AND is less than or equal to 3

Position function:
$\displaystyle f(x) = 2x^3 - 6x^2 + 4x + 5$

Velocity:

$\displaystyle f'(x) = 6x^2 - 12x + 4$

Acceleration:

$\displaystyle f''(x) = 12x - 12$

I'm asked to find when the object is stationary. Now correct me if I'm wrong but that should be when the slopes of the position function = 0 or rather the x-intercepts of the velocity function.

However i'm having trouble finding the x-intercepts

2. Find the equation of a normal line (perpendicular to a tangent line) of
$\displaystyle y = \sqrt{x} + 1/\sqrt[3]{x}$ at x = 1

What I did was solve for y which is y=2

So I have a point now (1,2)

Then I find the derivative of the function which is
y' = (1/2)(x^(-1/2)) + (-1/3)(x^(-4/3))

I then plug in 1 which is y' = 1/6 at x =1

I use the point slope equation

y - 2 = 1/6(x-1)
y = 1/6x -1/6 + 2
y = 1/6x + 11/6

Since the equation of a normal line is the negative reciprocal:
y = -1/(1/6x + 11/6)

Is this right or not I just want to confirm

2. Originally Posted by GameTheory
1.Suppose that "f" is an objects position at time "x" in seconds.

x is greater than or equal to 0 AND is less than or equal to 3

Position function:
$\displaystyle f(x) = 2x^3 - 6x^2 + 4x + 5$

Velocity:

$\displaystyle f'(x) = 6x^2 - 12x + 4$

Acceleration:

$\displaystyle f"(x) = 12x - 12$

I'm asked to find when the object is stationary. Now correct me if I'm wrong but that should be when the slopes of the position function = 0 or rather the x-intercepts of the velocity function.

However i'm having trouble finding the x-intercepts
Your derivatives are correct. The object is stationary when the velocity $\displaystyle f'(x) = 0$.

So $\displaystyle 6x^2 - 12x + 4 = 0$

$\displaystyle x^2 - 2x + \frac{2}{3} = 0$

$\displaystyle x^2 - 2x + (-1)^2 - (-1)^2 + \frac{2}{3} = 0$

$\displaystyle (x - 1)^2 - 1 + \frac{2}{3} = 0$

$\displaystyle (x - 1)^2 - \frac{1}{3} = 0$

$\displaystyle (x - 1)^2 = \frac{1}{3}$

$\displaystyle x - 1 = \pm \sqrt{\frac{1}{3}}$

$\displaystyle x - \frac{3}{3} = \frac{\pm \sqrt{3}}{3}$

$\displaystyle x = \frac{3 \pm \sqrt{3}}{3}$.

So the object is stationary when $\displaystyle x = \frac{3 - \sqrt{3}}{3}$ and when $\displaystyle x = \frac{3 + \sqrt{3}}{3}$.

3. Originally Posted by GameTheory
2. Find the equation of a normal line (perpendicular to a tangent line) of
$\displaystyle y = \sqrt{x} + 1/\sqrt[3]{x}$ at x = 1

What I did was solve for y which is y=2

So I have a point now (1,2)

Then I find the derivative of the function which is
y' = (1/2)(x^(-1/2)) + (-1/3)(x^(-4/3))

I then plug in 1 which is y' = 1/6 at x =1

I use the point slope equation

y - 2 = 1/6(x-1)
y = 1/6x -1/6 + 2
y = 1/6x + 11/6

Since the equation of a normal line is the negative reciprocal:
y = -1/(1/6x + 11/6)

Is this right or not I just want to confirm

Your evaluation of $\displaystyle y$ is correct. So $\displaystyle (1, 2)$ lies on the normal line.

If $\displaystyle m_T$ is the gradient of line tangent to the curve at $\displaystyle (x, y) = (1, 2)$,

then $\displaystyle m_N = -\frac{1}{m_T}$.

So to find $\displaystyle m_T$...

$\displaystyle y = \sqrt{x} + \frac{1}{\sqrt[3]{x}}$

$\displaystyle = x^{\frac{1}{2}} + x^{-\frac{1}{3}}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{3}x^{-\frac{4}{3}}$

$\displaystyle = \frac{1}{2\sqrt{x}} - \frac{1}{3\sqrt[3]{x^4}}$

$\displaystyle m_T = \frac{dy}{dx}|_{x = 1}$

$\displaystyle = \frac{1}{2\sqrt{1}} - \frac{1}{3\sqrt{1^4}}$

$\displaystyle = \frac{1}{2} - \frac{1}{3}$

$\displaystyle = \frac{1}{6}$.

Therefore $\displaystyle m_N = -\frac{1}{m_T}$

$\displaystyle = -\frac{1}{\frac{1}{6}}$

$\displaystyle = -6$.

So now you have the gradient of the normal line.

The normal line is of the form

$\displaystyle y = mx + c$

Substituting $\displaystyle (x, y) = (1, 2)$ and $\displaystyle m = -6$ gives

$\displaystyle 2 = -6(1) + c$

$\displaystyle 2 = -6 + c$

$\displaystyle c = 8$.

Therefore, the equation of the normal line is

$\displaystyle y = -6x + 8$.

4. Thanks alot, I realize now where I made my mistake in no.2.
However I would like to know what this process is:
$\displaystyle (-1)^2 - (-1)^2$

of

$\displaystyle x^2 - 2x + (-1)^2 - (-1)^2 + \frac{2}{3} = 0$

I don't understand how you just plugged in those -1s

5. Originally Posted by GameTheory
Thanks alot, I realize now where I made my mistake in no.2.
However I would like to know what this process is:
$\displaystyle (-1)^2 - (-1)^2$

of

$\displaystyle x^2 - 2x + (-1)^2 - (-1)^2 + \frac{2}{3} = 0$

I don't understand how you just plugged in those -1s
Completing the square.

6. 1.Suppose that "f" is an objects position at time "x" in seconds.

x is greater than or equal to 0 AND is less than or equal to 3

Position function:
$\displaystyle f(x) = 2x^3 - 6x^2 + 4x + 5$

Velocity:

$\displaystyle f'(x) = 6x^2 - 12x + 4$

Acceleration:

$\displaystyle f''(x) = 12x - 12$

I am also asked to find when the object is moving the fastest. How would I go about in doing this?

7. Originally Posted by GameTheory
1.Suppose that "f" is an objects position at time "x" in seconds.

x is greater than or equal to 0 AND is less than or equal to 3

Position function:
$\displaystyle f(x) = 2x^3 - 6x^2 + 4x + 5$

Velocity:

$\displaystyle f'(x) = 6x^2 - 12x + 4$

Acceleration:

$\displaystyle f''(x) = 12x - 12$

I am also asked to find when the object is moving the fastest. How would I go about in doing this?
The object is moving the fastest when the velocity is maximised.

You have a velocity function $\displaystyle f'(x)$. How would you find the point where it is maximised?

8. x interval is between 0 and 3. So I suppose f'(3) = maximized velocity?

so f'(3) = 22 so the object is moving fastest at x = 3?