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Finding inflection points- where did I miscalculate?
Clash Royale CLAN TAG#URR8PPP
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$3 sin x- sin^3x$ on $[0, 2pi]$
First derivative is :$3 cos x-3 sin^2x cos x$.
I have written it like this: $3 cos x(1-sin^2x)$.
Using the identity $1 - sin^2x = cos^2x$, I get: $3 cos x cos^2x$.
I mark '$cos x$' as 't' and then: second derivative is $9t^2$.
Comparing it to $0$, I get $x=pi/2$ and $3pi/2$.
In the workbook, it didn't assign value $t$ as $cos x$ but instead went like this:
$-9 cos^2xsin x$
That is the second derivative and the span of solution it gets is: $x=0,x=pi/2,x=pi$ and $x=3pi/2$. I fully understand this way, but why does marking '$cos x$' as '$t$' get me a different span of solutions than in the workbook?
calculus trigonometry
edited Jul 18 at 15:18
N. F. Taussig
38.2k93053
asked Jul 18 at 11:59
user6394019
30311
add a comment |Â
up vote
0
down vote
favorite
$3 sin x- sin^3x$ on $[0, 2pi]$
First derivative is :$3 cos x-3 sin^2x cos x$.
I have written it like this: $3 cos x(1-sin^2x)$.
Using the identity $1 - sin^2x = cos^2x$, I get: $3 cos x cos^2x$.
I mark '$cos x$' as 't' and then: second derivative is $9t^2$.
Comparing it to $0$, I get $x=pi/2$ and $3pi/2$.
In the workbook, it didn't assign value $t$ as $cos x$ but instead went like this:
$-9 cos^2xsin x$
That is the second derivative and the span of solution it gets is: $x=0,x=pi/2,x=pi$ and $x=3pi/2$. I fully understand this way, but why does marking '$cos x$' as '$t$' get me a different span of solutions than in the workbook?
calculus trigonometry
edited Jul 18 at 15:18
N. F. Taussig
38.2k93053
asked Jul 18 at 11:59
user6394019
30311
1
$fracd f(t)dx=fracd f(t)dt fracdtdx$
– random
Jul 18 at 12:38
@random I didn't grasp the idea. can you please show the equations?
– user6394019
Jul 18 at 12:51
1
@use6394019 The idea is that by definition you need the second derivative with respect to $x$ and that there is no reason for thinking that the derivative with respect to $t$ will have the same zeroes. By ignoring the second part of the chain rule you miss the zeroes of $fracd cos xdx$.
– random
Jul 18 at 13:15
The second derivative does not change sign at $x = 0$. The inflection points are $x = fracpi2$, $x = pi$, and $x = frac3pi2$.
– N. F. Taussig
Jul 18 at 15:19
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$3 sin x- sin^3x$ on $[0, 2pi]$
First derivative is :$3 cos x-3 sin^2x cos x$.
I have written it like this: $3 cos x(1-sin^2x)$.
Using the identity $1 - sin^2x = cos^2x$, I get: $3 cos x cos^2x$.
I mark '$cos x$' as 't' and then: second derivative is $9t^2$.
Comparing it to $0$, I get $x=pi/2$ and $3pi/2$.
In the workbook, it didn't assign value $t$ as $cos x$ but instead went like this:
$-9 cos^2xsin x$
That is the second derivative and the span of solution it gets is: $x=0,x=pi/2,x=pi$ and $x=3pi/2$. I fully understand this way, but why does marking '$cos x$' as '$t$' get me a different span of solutions than in the workbook?
calculus trigonometry
edited Jul 18 at 15:18
N. F. Taussig
38.2k93053
asked Jul 18 at 11:59
user6394019
30311
$3 sin x- sin^3x$ on $[0, 2pi]$
First derivative is :$3 cos x-3 sin^2x cos x$.
I have written it like this: $3 cos x(1-sin^2x)$.
Using the identity $1 - sin^2x = cos^2x$, I get: $3 cos x cos^2x$.
I mark '$cos x$' as 't' and then: second derivative is $9t^2$.
Comparing it to $0$, I get $x=pi/2$ and $3pi/2$.
In the workbook, it didn't assign value $t$ as $cos x$ but instead went like this:
$-9 cos^2xsin x$
That is the second derivative and the span of solution it gets is: $x=0,x=pi/2,x=pi$ and $x=3pi/2$. I fully understand this way, but why does marking '$cos x$' as '$t$' get me a different span of solutions than in the workbook?
calculus trigonometry
edited Jul 18 at 15:18
N. F. Taussig
38.2k93053
asked Jul 18 at 11:59
user6394019
30311
edited Jul 18 at 15:18
N. F. Taussig
38.2k93053
edited Jul 18 at 15:18
N. F. Taussig
38.2k93053
edited Jul 18 at 15:18
N. F. Taussig
38.2k93053
38.2k93053
asked Jul 18 at 11:59
user6394019
30311
asked Jul 18 at 11:59
user6394019
30311
asked Jul 18 at 11:59
user6394019
30311
30311
1
$fracd f(t)dx=fracd f(t)dt fracdtdx$
– random
Jul 18 at 12:38
@random I didn't grasp the idea. can you please show the equations?
– user6394019
Jul 18 at 12:51
1
@use6394019 The idea is that by definition you need the second derivative with respect to $x$ and that there is no reason for thinking that the derivative with respect to $t$ will have the same zeroes. By ignoring the second part of the chain rule you miss the zeroes of $fracd cos xdx$.
– random
Jul 18 at 13:15
The second derivative does not change sign at $x = 0$. The inflection points are $x = fracpi2$, $x = pi$, and $x = frac3pi2$.
– N. F. Taussig
Jul 18 at 15:19
add a comment |Â
1
$fracd f(t)dx=fracd f(t)dt fracdtdx$
– random
Jul 18 at 12:38
@random I didn't grasp the idea. can you please show the equations?
– user6394019
Jul 18 at 12:51
1
@use6394019 The idea is that by definition you need the second derivative with respect to $x$ and that there is no reason for thinking that the derivative with respect to $t$ will have the same zeroes. By ignoring the second part of the chain rule you miss the zeroes of $fracd cos xdx$.
– random
Jul 18 at 13:15
The second derivative does not change sign at $x = 0$. The inflection points are $x = fracpi2$, $x = pi$, and $x = frac3pi2$.
– N. F. Taussig
Jul 18 at 15:19
1
1
$fracd f(t)dx=fracd f(t)dt fracdtdx$
– random
Jul 18 at 12:38
$fracd f(t)dx=fracd f(t)dt fracdtdx$
– random
Jul 18 at 12:38
@random I didn't grasp the idea. can you please show the equations?
– user6394019
Jul 18 at 12:51
@random I didn't grasp the idea. can you please show the equations?
– user6394019
Jul 18 at 12:51
1
1
@use6394019 The idea is that by definition you need the second derivative with respect to $x$ and that there is no reason for thinking that the derivative with respect to $t$ will have the same zeroes. By ignoring the second part of the chain rule you miss the zeroes of $fracd cos xdx$.
– random
Jul 18 at 13:15
@use6394019 The idea is that by definition you need the second derivative with respect to $x$ and that there is no reason for thinking that the derivative with respect to $t$ will have the same zeroes. By ignoring the second part of the chain rule you miss the zeroes of $fracd cos xdx$.
– random
Jul 18 at 13:15
The second derivative does not change sign at $x = 0$. The inflection points are $x = fracpi2$, $x = pi$, and $x = frac3pi2$.
– N. F. Taussig
Jul 18 at 15:19
The second derivative does not change sign at $x = 0$. The inflection points are $x = fracpi2$, $x = pi$, and $x = frac3pi2$.
– N. F. Taussig
Jul 18 at 15:19
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
You failed to apply the Chain Rule.
Chain Rule. Let $h = g circ f$. Suppose that both the derivatives $f'(x)$ and $g'(y)$ exist, where $y = f(x)$. Then the derivative $h'(x)$ also exists and is given by the formula $h'(x) = f'(x) cdot g'(f(x))$.
You differentiated $y = 3sin x - sin^3x$ to obtain
beginalign*
y' & = 3cos x - 3sin^2xcos x\
& = 3cos x(1 - sin^2x)\
& = 3cos x(cos^2x)\
& = 3cos^3x
endalign*
Thus, the second derivative is
beginalign*
y'' & = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
You made the substitution $t = cos x$ to write $y'(t) = 3t^3$. However, observe that since $t = cos x$, what we really have is $y'(t(x)) = 3[t(x)]^3$. Applying the Chain Rule yields
beginalign*
y''(t(x)) & = 9[t(x)]^2[t'(x)]\
& = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
in agreement with the result we obtained above.
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
add a comment |Â
up vote
0
down vote
After you apply the clain rule to obtain the second derivative (see N. F. Taussig's description below), your derivative should have been $f''(t) = 9t^2 t'$.
Then, you can split the resulting equation $-9 cos^2 x sin x = 0$ into two parts to find the inflection points.
For your first part ($-9 cos^2 x = 0$) the two solutions you obtained are absolutely correct ($pi/2 text and 3pi/2$). For the second part ($sin x = 0$) we have $x = 0 text or x = pi$. Plugging in $x = 0$ gives us no change in sign, so no inflection point, but plugging in $pi$ does; hence the three inflection points are $pi/2, pi text , and 3pi/2.$
answered Jul 18 at 12:23
bjcolby15
8171716
The question is about the Chain Rule.
– N. F. Taussig
Jul 18 at 12:40
@bjcolby but why using my way I didnt get all the solutions, but only a part of it?
– user6394019
Jul 18 at 12:50
I've cleaned up my explanation above. I agree with N.F. Taussig's explanation; I was trying to explain the answers to the resulting trigonometry question (which I admit I didn't do too well :/ )
– bjcolby15
Jul 18 at 15:47
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You failed to apply the Chain Rule.
Chain Rule. Let $h = g circ f$. Suppose that both the derivatives $f'(x)$ and $g'(y)$ exist, where $y = f(x)$. Then the derivative $h'(x)$ also exists and is given by the formula $h'(x) = f'(x) cdot g'(f(x))$.
You differentiated $y = 3sin x - sin^3x$ to obtain
beginalign*
y' & = 3cos x - 3sin^2xcos x\
& = 3cos x(1 - sin^2x)\
& = 3cos x(cos^2x)\
& = 3cos^3x
endalign*
Thus, the second derivative is
beginalign*
y'' & = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
You made the substitution $t = cos x$ to write $y'(t) = 3t^3$. However, observe that since $t = cos x$, what we really have is $y'(t(x)) = 3[t(x)]^3$. Applying the Chain Rule yields
beginalign*
y''(t(x)) & = 9[t(x)]^2[t'(x)]\
& = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
in agreement with the result we obtained above.
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
add a comment |Â
up vote
3
down vote
You failed to apply the Chain Rule.
Chain Rule. Let $h = g circ f$. Suppose that both the derivatives $f'(x)$ and $g'(y)$ exist, where $y = f(x)$. Then the derivative $h'(x)$ also exists and is given by the formula $h'(x) = f'(x) cdot g'(f(x))$.
You differentiated $y = 3sin x - sin^3x$ to obtain
beginalign*
y' & = 3cos x - 3sin^2xcos x\
& = 3cos x(1 - sin^2x)\
& = 3cos x(cos^2x)\
& = 3cos^3x
endalign*
Thus, the second derivative is
beginalign*
y'' & = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
You made the substitution $t = cos x$ to write $y'(t) = 3t^3$. However, observe that since $t = cos x$, what we really have is $y'(t(x)) = 3[t(x)]^3$. Applying the Chain Rule yields
beginalign*
y''(t(x)) & = 9[t(x)]^2[t'(x)]\
& = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
in agreement with the result we obtained above.
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You failed to apply the Chain Rule.
Chain Rule. Let $h = g circ f$. Suppose that both the derivatives $f'(x)$ and $g'(y)$ exist, where $y = f(x)$. Then the derivative $h'(x)$ also exists and is given by the formula $h'(x) = f'(x) cdot g'(f(x))$.
You differentiated $y = 3sin x - sin^3x$ to obtain
beginalign*
y' & = 3cos x - 3sin^2xcos x\
& = 3cos x(1 - sin^2x)\
& = 3cos x(cos^2x)\
& = 3cos^3x
endalign*
Thus, the second derivative is
beginalign*
y'' & = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
You made the substitution $t = cos x$ to write $y'(t) = 3t^3$. However, observe that since $t = cos x$, what we really have is $y'(t(x)) = 3[t(x)]^3$. Applying the Chain Rule yields
beginalign*
y''(t(x)) & = 9[t(x)]^2[t'(x)]\
& = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
in agreement with the result we obtained above.
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
You failed to apply the Chain Rule.
Chain Rule. Let $h = g circ f$. Suppose that both the derivatives $f'(x)$ and $g'(y)$ exist, where $y = f(x)$. Then the derivative $h'(x)$ also exists and is given by the formula $h'(x) = f'(x) cdot g'(f(x))$.
You differentiated $y = 3sin x - sin^3x$ to obtain
beginalign*
y' & = 3cos x - 3sin^2xcos x\
& = 3cos x(1 - sin^2x)\
& = 3cos x(cos^2x)\
& = 3cos^3x
endalign*
Thus, the second derivative is
beginalign*
y'' & = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
You made the substitution $t = cos x$ to write $y'(t) = 3t^3$. However, observe that since $t = cos x$, what we really have is $y'(t(x)) = 3[t(x)]^3$. Applying the Chain Rule yields
beginalign*
y''(t(x)) & = 9[t(x)]^2[t'(x)]\
& = 9cos^2x(-sin x)\
& = -9cos^2xsin x
endalign*
in agreement with the result we obtained above.
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
answered Jul 18 at 15:15
N. F. Taussig
38.2k93053
38.2k93053
add a comment |Â
add a comment |Â
up vote
0
down vote
After you apply the clain rule to obtain the second derivative (see N. F. Taussig's description below), your derivative should have been $f''(t) = 9t^2 t'$.
Then, you can split the resulting equation $-9 cos^2 x sin x = 0$ into two parts to find the inflection points.
For your first part ($-9 cos^2 x = 0$) the two solutions you obtained are absolutely correct ($pi/2 text and 3pi/2$). For the second part ($sin x = 0$) we have $x = 0 text or x = pi$. Plugging in $x = 0$ gives us no change in sign, so no inflection point, but plugging in $pi$ does; hence the three inflection points are $pi/2, pi text , and 3pi/2.$
answered Jul 18 at 12:23
bjcolby15
8171716
The question is about the Chain Rule.
– N. F. Taussig
Jul 18 at 12:40
@bjcolby but why using my way I didnt get all the solutions, but only a part of it?
– user6394019
Jul 18 at 12:50
I've cleaned up my explanation above. I agree with N.F. Taussig's explanation; I was trying to explain the answers to the resulting trigonometry question (which I admit I didn't do too well :/ )
– bjcolby15
Jul 18 at 15:47
add a comment |Â
up vote
0
down vote
After you apply the clain rule to obtain the second derivative (see N. F. Taussig's description below), your derivative should have been $f''(t) = 9t^2 t'$.
Then, you can split the resulting equation $-9 cos^2 x sin x = 0$ into two parts to find the inflection points.
For your first part ($-9 cos^2 x = 0$) the two solutions you obtained are absolutely correct ($pi/2 text and 3pi/2$). For the second part ($sin x = 0$) we have $x = 0 text or x = pi$. Plugging in $x = 0$ gives us no change in sign, so no inflection point, but plugging in $pi$ does; hence the three inflection points are $pi/2, pi text , and 3pi/2.$
answered Jul 18 at 12:23
bjcolby15
8171716
The question is about the Chain Rule.
– N. F. Taussig
Jul 18 at 12:40
@bjcolby but why using my way I didnt get all the solutions, but only a part of it?
– user6394019
Jul 18 at 12:50
I've cleaned up my explanation above. I agree with N.F. Taussig's explanation; I was trying to explain the answers to the resulting trigonometry question (which I admit I didn't do too well :/ )
– bjcolby15
Jul 18 at 15:47
add a comment |Â
up vote
0
down vote
up vote
0
down vote
After you apply the clain rule to obtain the second derivative (see N. F. Taussig's description below), your derivative should have been $f''(t) = 9t^2 t'$.
Then, you can split the resulting equation $-9 cos^2 x sin x = 0$ into two parts to find the inflection points.
For your first part ($-9 cos^2 x = 0$) the two solutions you obtained are absolutely correct ($pi/2 text and 3pi/2$). For the second part ($sin x = 0$) we have $x = 0 text or x = pi$. Plugging in $x = 0$ gives us no change in sign, so no inflection point, but plugging in $pi$ does; hence the three inflection points are $pi/2, pi text , and 3pi/2.$
answered Jul 18 at 12:23
bjcolby15
8171716
After you apply the clain rule to obtain the second derivative (see N. F. Taussig's description below), your derivative should have been $f''(t) = 9t^2 t'$.
Then, you can split the resulting equation $-9 cos^2 x sin x = 0$ into two parts to find the inflection points.
For your first part ($-9 cos^2 x = 0$) the two solutions you obtained are absolutely correct ($pi/2 text and 3pi/2$). For the second part ($sin x = 0$) we have $x = 0 text or x = pi$. Plugging in $x = 0$ gives us no change in sign, so no inflection point, but plugging in $pi$ does; hence the three inflection points are $pi/2, pi text , and 3pi/2.$
answered Jul 18 at 12:23
bjcolby15
8171716
edited Jul 18 at 15:45
answered Jul 18 at 12:23
bjcolby15
8171716
answered Jul 18 at 12:23
bjcolby15
8171716
answered Jul 18 at 12:23
bjcolby15
8171716
8171716
The question is about the Chain Rule.
– N. F. Taussig
Jul 18 at 12:40
@bjcolby but why using my way I didnt get all the solutions, but only a part of it?
– user6394019
Jul 18 at 12:50
I've cleaned up my explanation above. I agree with N.F. Taussig's explanation; I was trying to explain the answers to the resulting trigonometry question (which I admit I didn't do too well :/ )
– bjcolby15
Jul 18 at 15:47
add a comment |Â
The question is about the Chain Rule.
– N. F. Taussig
Jul 18 at 12:40
@bjcolby but why using my way I didnt get all the solutions, but only a part of it?
– user6394019
Jul 18 at 12:50
I've cleaned up my explanation above. I agree with N.F. Taussig's explanation; I was trying to explain the answers to the resulting trigonometry question (which I admit I didn't do too well :/ )
– bjcolby15
Jul 18 at 15:47
The question is about the Chain Rule.
– N. F. Taussig
Jul 18 at 12:40
The question is about the Chain Rule.
– N. F. Taussig
Jul 18 at 12:40
@bjcolby but why using my way I didnt get all the solutions, but only a part of it?
– user6394019
Jul 18 at 12:50
@bjcolby but why using my way I didnt get all the solutions, but only a part of it?
– user6394019
Jul 18 at 12:50
I've cleaned up my explanation above. I agree with N.F. Taussig's explanation; I was trying to explain the answers to the resulting trigonometry question (which I admit I didn't do too well :/ )
– bjcolby15
Jul 18 at 15:47
I've cleaned up my explanation above. I agree with N.F. Taussig's explanation; I was trying to explain the answers to the resulting trigonometry question (which I admit I didn't do too well :/ )
– bjcolby15
Jul 18 at 15:47
add a comment |Â
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1
$fracd f(t)dx=fracd f(t)dt fracdtdx$
– random
Jul 18 at 12:38
@random I didn't grasp the idea. can you please show the equations?
– user6394019
Jul 18 at 12:51
1
@use6394019 The idea is that by definition you need the second derivative with respect to $x$ and that there is no reason for thinking that the derivative with respect to $t$ will have the same zeroes. By ignoring the second part of the chain rule you miss the zeroes of $fracd cos xdx$.
– random
Jul 18 at 13:15
The second derivative does not change sign at $x = 0$. The inflection points are $x = fracpi2$, $x = pi$, and $x = frac3pi2$.
– N. F. Taussig
Jul 18 at 15:19