Derivative problems

1. Oct 10, 2005

seiferseph

1) i'm having trouble with this problem, it looks like their isn't enough information. can someone give me some hints please
http://i2.photobucket.com/albums/y15/seiferseph/untitled.jpg [Broken]

2)
http://i2.photobucket.com/albums/y15/seiferseph/untitled2.jpg [Broken]
i solved parts a-c fine, but i'm having some trouble with d and e
i solved v(t) 6t^2 - 30t + 2 and a(t) = 12t - 30 (taking the derivative). for d and e it is looking for a range, right? heres what i got

d) it is moving left when v(t) < 0, so between the zeros, or 0.676 < t < 4.93
and it is moving right when v(t) > 0, or t < 0.676 and t>2.5

e) p is speeding up when both v(t) and a(t) are in the same direction, and slowing down when they are opposite directions (signs)
i got a(t) > 0 when t > 5/2, and a(t) < 0 when t < 5/2. so its speeding up when t > 4.93 and slowing when 0.676 < t < 2.5

Last edited by a moderator: May 2, 2017
2. Oct 10, 2005

Fermat

For 1),

y = x²
y' = 2x

From this you can get the slope of the tangent at A(1,1) , hence also the slope of the normal. You know the coordinates of a point and the slope of the line passing through that point, A, hence you can find the eqn of that line.

Will that get you started ?

3. Oct 10, 2005

Fermat

4. Oct 10, 2005

seiferseph

yes that helped, thanks! i guess i just saw it and got confused :surprised

so heres what i did, i sovled the slope of the normal line to be -1/2, which gave it the equation y = -1/2x + 3/2; i equated this to y=x^2 to solve for the intersection points, which i got to be (1, 1) and (-3/2, 9/4). i solved for the tangent line for x = -3/2 to get the slope to be -3, and i get the line equation to be y = -3x - 9/4. is this correct?

btw, thanks for confirming the other one too! i just worked my way through it, i haven't seen a problem like that before

5. Oct 10, 2005

Fermat

I get the same as yourself for all the results you gave

6. Oct 10, 2005

seiferseph

thanks again, but i was thinking about part e of that second question, and when is it actually slowing down? would it be when they have the opposite sign? how does that work?

7. Oct 10, 2005

Fermat

It is slowing down when velocity and acceleration have opposite signs.

Consider the eqn of motion,

vf = vi + at

where vf is the final velocity after time t, and vi is some inital velocity.

Let vf be +ve (positive) and let a be -ve (negative).
If a is -ve, then vi +at will be continually getting smaller as t increases, hence vf (= vi + at) will be decreasing. So the speed is getting (numerically) smaller.

Now, let vf be -ve and let a be +ve.
If a is +ve, then vi +at will be continually getting larger as t increases, hence vf (= vi + at) will be increasing.
But vf is -ve. And when -ve numbers increase they get numerically smaller. i.e the speed is getting smaller.

8. Oct 10, 2005

seiferseph

so for this problem would it be

speeding up when t>4.93 and when 0.676 < t < 2.5 ? and slowing down when they are opposite signs, at 2.5 < t < 4.93 and t < 0.676? i don't completely understand when it would be speeding or slowing in this problem...