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:
Velocity:
Acceleration:
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
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
Thanks in advance
Your evaluation of is correct. So lies on the normal line.
Now you need the gradient.
If is the gradient of line tangent to the curve at ,
then .
So to find ...
.
Therefore
.
So now you have the gradient of the normal line.
The normal line is of the form
Substituting and gives
.
Therefore, the equation of the normal line is
.
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:
Velocity:
Acceleration:
I am also asked to find when the object is moving the fastest. How would I go about in doing this?