calculus problem... im stuck

[schneiderguy]

Here's the question:

"The position of a particle moving along a line is given by s(t) = 2t^3 - 24t^2 + 90t + 7 for t >= 0. For what values of t is the speed of the particle increasing?"

The second derivative of a position function is the acceleration, right? So when the second derivative is positive (positive acceleration) that is where the speed of the particle would be increasing.

s'(t) = 6t^2 - 48t + 90
s''(t) = 12t - 48

From that I get that for anything over t=4 the acceleration is positive thus the speed is increasing... but the correct answer should be 3 < t < 4 && t > 5 according to the answer key.

What am I doing wrong? Any help would be appreciated

 

[HammerCurl]

What you said sound right, I wouldnt worry to much about it. Ask your TA or Prof if you're really that concerned, but answer keys aren't always perfect. The answer in the key would be right if it was asking for "For what values of t is the speed of the particle positive."

 

[Gibson486]

graph the first derivative.

edit: I just graphed it, I think somethig is wrong.....


BTW: double derivatives will only show point of inflection. You have to do further tests to prove whether it is negative or positive acceleration.

 

[MrChad]

Is the answer key part of a text book? Have you checked online to see if errata are posted?

 

[amoeba]

schneider, I think its a misprint in either the question or the answer key. Go with what HammerCurl suggested.

 

[schneiderguy]

Originally posted by: MrChad
Is the answer key part of a text book? Have you checked online to see if errata are posted?

It is part of the AP Calculus practice multiple choice questions here. It is #9

Thanks everyone

 

[Itchrelief]

I think they are getting cute with speed vs. velocity. Btwn t=3 and t=4, velocity is decreasing, but speed is increasing, since speed is the absolute value of velocity.

 

[ChAoTiCpInOy]

Don't you just graph the first derivative and see where it is above the x-axis?

 

[StevenYoo]

Originally posted by: Itchrelief
I think they are getting cute with speed vs. velocity. Btwn t=3 and t=4, velocity is decreasing, but speed is increasing, since speed is the absolute value of velocity.

if that's true, then that's a bitch of a question testing your knowledge of semantics as well a calculus.

However, if that were indeed the case, then at any point where you have a non-zero slope, speed would technically be increasing, be it positively or negatively. That can't be it.

 

[Itchrelief]

Originally posted by: StevenYoo

if that's true, then that's a bitch of a question testing your knowledge of semantics as well a calculus.

However, if that were indeed the case, then at any point where you have a non-zero slope, speed would technically be increasing, be it positively or negatively. That can't be it.

I agree that it's a semantics question, but not that any nonzero slope (I assume you mean acceleration) means speed increases.

Speed increases when the magnitude of the velocity increases.

This is when velocity and acceleration are in the same direction.

Speed can decrease with nonzero acceleration if it is in the opposite direction to the velocity (at t<3 and 4<t<5). Funky behavior that occurs when the velocity graph drops below zero. <-- this last part is probably not technically correct, but that's one of the situations it tends to crop up, in my experience

 

[DrPizza]

Positive acceleration does not mean that you're speed is increasing; only that the velocity is increasing. You could have positive acceleration, but decreasing speed.

The easy way to remember it:
If the sign of acceleration and velocity are the same, then the object is speeding up. If the sign of the acceleration and the velocity are different, then the speed is decreasing.

So, you can't just look at acceleration alone. Nor can you look at velocity alone when you're analyzing it this way. To understand why the speed decreases, imagine a car on a very slight hill. And, let's define uphill as the positive direction; downhill as the negative direction. So, whenever the car is moving uphill, it's velocity is positive. Whenever the car is moving downhill, the car has a negative velocity.

Now, if the car is moving backwards down the hill, you can apply just the right amount of braking such that the speed stays the same. i.e. velocity is negative but the acceleration is zero.

Or, let the car roll backwards down the hill without the brakes on. Acceleration is negative, velocity is negative, but the car is speeding up!

Now, when the car is going backwards at about 10mph, (it's a standard transmission, btw), slowly let out the clutch, (thankfully, Calculus teachers don't have to really do this demonstration, else I'd be replacing a clutch every year!) This will cause the car to slow down. The clutch is allowing the car to accelerate in the uphill direction. It'll take a minute for 1st gear to bring the car to a momentary stop, then continue accelerating the car as its direction switches to uphill. This is negative velocity with a positive acceleration: the car slows down. And, in this case, I continued the same acceleration until the cars speed reached zero (zero speed/velocity), then as the car continued with that acceleration, the velocity became positive.

Which leads us to: If you're heading uphill and you press down on the gas (and the gears are engaged, of course), then you have positive velocity, positive acceleration, and you're speeding up.

And lastly, if the car is going uphill, and you put it in neutral, then the hill is going to cause acceleration toward the downhill side. i.e. negative acceleration while you have a positive velocity. The car is slowing down. (And will eventually slow down to zero velocity, then the velocity will become negative and the car will speed up with negative velocity & negative acceleration.



Hope this helped!

 

[DrPizza]

Oh, and it's not semantics. It's common sense. Well, uncommon sense; it's common math-sense after you've mastered this little portion of calculus
And, this helps reinforce my view that those damn calculators led to less understanding of calculus for those of you who have taken calculus already.


To answer your question according to what I have above:

s'(t) = velocity = 0 at
t²-8t+15=0 (sorry, had to think out loud)

= 0 at 3 and 5 seconds. Since it's a "happy parabola", obviously the graph of velocity would show that it's negative between 3 and 5 seconds; positive at t<3 seconds, and positive at t>5 seconds.


Acceleration is 0 when t = 4 Since the graph of acceleration is a line with a positive slope, acceleration is negative when t<4 and positive when t>4

So, from 0 to 3 seconds (assuming the stopwatch starts at zero), acceleration is negative and velocity is positive. Opposite signs; it's slowing down.

From t=3 seconds to t=4 seconds, acceleration is negative and velocity is negative. Signs are the same, so it's speeding up.

From t = 4 seconds to t=5 seconds, acceleration is positive and velocity is still negative. Signs are different, so it's slowing down.

From t = 5 seconds on, acceleration is positive and velocity is positive; signs are the same, it's speeding up.

Observing an object, you'd see it starting with a speed of 90 going uphill, and slowing down (engine is turned off as it's going up a hill. At the 3 second point, it's momentarily stopped, then it starts going backwards (speeding up downhill). At the 4 second point, the engine is turned on and the car starts putting out power to make it go uphill. But, it isn't until the 5 second mark that the car comes to rest momentarily again and reverses its direction.


(edit, if someone knows of a better analogy than the car on a hill, please PM me with it. I've been using it in both calculus and physics with great success, but I'm always looking to improve my teaching.)

 

[Gibson486]

Originally posted by: DrPizza
Oh, and it's not semantics. It's common sense. Well, uncommon sense; it's common math-sense after you've mastered this little portion of calculus
And, this helps reinforce my view that those damn calculators led to less understanding of calculus for those of you who have taken calculus already.


To answer your question according to what I have above:

s'(t) = velocity = 0 at
t²-8t+15=0 (sorry, had to think out loud)

= 0 at 3 and 5 seconds. Since it's a "happy parabola", obviously the graph of velocity would show that it's negative between 3 and 5 seconds; positive at t<3 seconds, and positive at t>5 seconds.


Acceleration is 0 when t = 4 Since the graph of acceleration is a line with a positive slope, acceleration is negative when t<4 and positive when t>4

So, from 0 to 3 seconds (assuming the stopwatch starts at zero), acceleration is negative and velocity is positive. Opposite signs; it's slowing down.

From t=3 seconds to t=4 seconds, acceleration is negative and velocity is negative. Signs are the same, so it's speeding up.

From t = 4 seconds to t=5 seconds, acceleration is positive and velocity is still negative. Signs are different, so it's slowing down.

From t = 5 seconds on, acceleration is positive and velocity is positive; signs are the same, it's speeding up.

Observing an object, you'd see it starting with a speed of 90 going uphill, and slowing down (engine is turned off as it's going up a hill. At the 3 second point, it's momentarily stopped, then it starts going backwards (speeding up downhill). At the 4 second point, the engine is turned on and the car starts putting out power to make it go uphill. But, it isn't until the 5 second mark that the car comes to rest momentarily again and reverses its direction.


(edit, if someone knows of a better analogy than the car on a hill, please PM me with it. I've been using it in both calculus and physics with great success, but I'm always looking to improve my teaching.)

hehehe.....I like how you say "it's common math-sense after you've mastered this little portion of calculus ".... so true....