Expert: Paul Klarreich - 2/8/2009

Question
QUESTION: Let f b the function defined for pi/6 ≤ x ≤ 5pi/6 by f(x)=x+ (sin x)^2.
c.) Find all the x-coordinates of all inflection points of f. Justify your answer

ANSWER: Questioner:   Linrosa
Category:  Calculus
Private:  No
 
Subject:  Inflection Points
Question:  Let f b the function defined for pi/6 ≤ x ≤ 5pi/6 by f(x)=x+ (sin x)^2.
c.) Find all the x-coordinates of all inflection points of f. Justify your answer
................................

since the second derivative of x is 0, you won't have any trouble.

f' = 1 + 2 sin x cos x = 1 + sin 2x

f'' = 2 cos 2x  

Set that = 0 and solve.  You will get:

2x = pi/2,  2x = 3pi/2

x = pi/4, x = 3pi/4.

That should do it.


---------- FOLLOW-UP ----------

QUESTION: Thanks so much for your help! I was wondering if you can help me with one more question for the day & I wont bother you anymore today:

A particle moves along the X-axis so that at time t its position is given by x(t)=sin(pi*t^2) for -1 ≤ t ≤ 1
a.) find the velocty at time t
b.) find the acceleration at time t
c.) for what value of t does the particle change direction?
d.) Find all values of t for which the particle is moving to the left

Answer
Questioner:   Linrosa
Category:  Calculus
Private:  No
 
---------- FOLLOW-UP ----------

QUESTION: Thanks so much for your help! I was wondering if you can help me with one more question for the day & I wont bother you anymore today:

This is quite standard stuff, and I think you really know how to do it:

A particle moves along the X-axis so that at time t its position is given by x(t)=sin(pi*t^2) for -1 ≤ t ≤ 1

a.) find the velocity at time t

This is just x'(t) = 2 pi t cos(pi t^2)

b.) find the acceleration at time t

This is just x''(t).  differentiate again. (product rule)

c.) for what value of t does the particle change direction?

This asks when x'(t) = 0.  

set t = 0,(one solution) and cos(pi t^2) = 0.  You should have pi t^2 = +- pi/2

So  t^2 = 1/2,  t = +- sqrt(2)/2

so  t = 0, +- sqrt(2)/2  are your solutions.

d.) Find all values of t for which the particle is moving to the left

This asks when (an interval)  x'(t) < 0:

x'(t) = 2 pi t cos(pi t^2)

This is negative when 't' and cos(pi t^2) have opposite signs.  
.....................
If t > 0, then look for  cos(pi t^2) to be negative.  

That means pi t^2 > pi/2, or t^2 > 1/2, or t > sqrt(2)/2.

Interval:  [sqrt(2)/2..1]
...................
If t < 0, then look for  cos(pi t^2) to be positive.  

That means pi t^2 < pi/2, or t^2 < 1/2, or t > - sqrt(2)/2.

(remember, in this little piece,  t is negative.)

Interval:  [- sqrt(2)/2..0]

That's it.  There is some tricky positive-negative analysis.

[See attached picture.]