Mixing
Second derivative of parametric equations
$begingroup$
I am looking for an intuitive explanation for the formula used to take the second derivative of a parametric function. The formula is:
$frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}$
I understand the reasoning for getting $frac{dy}{dx}$-- by dividing $frac{dy}{dt}$ by $frac{dx}{dt}$ -- however I am lost in the above formula. If we are finding the second derivative of y with respect to x, why do we differentiate $frac{dy}{dx}$ with respect to t, but not $frac{dx}{dt}$?
My inability to understand this could be due to my already fragile understanding of Leibnitz notation. At any rate, I would appreciate an explanation for why this formula is set up in this way.
Again, the $frac{dy}{dx}$ part makes sense, we were finding the derivative of y with respect to x, and since both x and y are functions of t, we found how x changes with respect to t, and how y changes with respect to t, and divided them.
Thanks in advance for any help!
calculus derivatives parametric
asked Jan 6 at 17:49
AddisonAddison
314
$endgroup$
add a comment |
$begingroup$
I am looking for an intuitive explanation for the formula used to take the second derivative of a parametric function. The formula is:
$frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}$
I understand the reasoning for getting $frac{dy}{dx}$-- by dividing $frac{dy}{dt}$ by $frac{dx}{dt}$ -- however I am lost in the above formula. If we are finding the second derivative of y with respect to x, why do we differentiate $frac{dy}{dx}$ with respect to t, but not $frac{dx}{dt}$?
My inability to understand this could be due to my already fragile understanding of Leibnitz notation. At any rate, I would appreciate an explanation for why this formula is set up in this way.
Again, the $frac{dy}{dx}$ part makes sense, we were finding the derivative of y with respect to x, and since both x and y are functions of t, we found how x changes with respect to t, and how y changes with respect to t, and divided them.
Thanks in advance for any help!
calculus derivatives parametric
asked Jan 6 at 17:49
AddisonAddison
314
$endgroup$
add a comment |
$begingroup$
I am looking for an intuitive explanation for the formula used to take the second derivative of a parametric function. The formula is:
$frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}$
I understand the reasoning for getting $frac{dy}{dx}$-- by dividing $frac{dy}{dt}$ by $frac{dx}{dt}$ -- however I am lost in the above formula. If we are finding the second derivative of y with respect to x, why do we differentiate $frac{dy}{dx}$ with respect to t, but not $frac{dx}{dt}$?
My inability to understand this could be due to my already fragile understanding of Leibnitz notation. At any rate, I would appreciate an explanation for why this formula is set up in this way.
Again, the $frac{dy}{dx}$ part makes sense, we were finding the derivative of y with respect to x, and since both x and y are functions of t, we found how x changes with respect to t, and how y changes with respect to t, and divided them.
Thanks in advance for any help!
calculus derivatives parametric
asked Jan 6 at 17:49
AddisonAddison
314
$endgroup$
I am looking for an intuitive explanation for the formula used to take the second derivative of a parametric function. The formula is:
$frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}$
I understand the reasoning for getting $frac{dy}{dx}$-- by dividing $frac{dy}{dt}$ by $frac{dx}{dt}$ -- however I am lost in the above formula. If we are finding the second derivative of y with respect to x, why do we differentiate $frac{dy}{dx}$ with respect to t, but not $frac{dx}{dt}$?
My inability to understand this could be due to my already fragile understanding of Leibnitz notation. At any rate, I would appreciate an explanation for why this formula is set up in this way.
Again, the $frac{dy}{dx}$ part makes sense, we were finding the derivative of y with respect to x, and since both x and y are functions of t, we found how x changes with respect to t, and how y changes with respect to t, and divided them.
Thanks in advance for any help!
calculus derivatives parametric
calculus derivatives parametric
asked Jan 6 at 17:49
AddisonAddison
314
asked Jan 6 at 17:49
AddisonAddison
314
asked Jan 6 at 17:49
AddisonAddison
314
asked Jan 6 at 17:49
AddisonAddison
314
asked Jan 6 at 17:49
AddisonAddison
314
314
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Intuitively (from Leibnitz notation):
$$
frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}=frac{d}{dt}left(frac{dy}{dx} right)frac{dt}{dx}
=
frac{d}{dx}left(frac{dy}{dx} right)
$$
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
$endgroup$
$begingroup$
Okay, so I see how the dt's cancel, creating the $frac{d}{dx}$. But why is the numerator so different than the denominator. When finding the first derivative, it is just $frac{frac{dy}{dt}}{frac{dx}{dt}}$, where the numerator and denominator are almost identical. Why is it that when you take the second derivative, the denominator looks so different from the numerator?
$endgroup$
– Addison
Jan 6 at 18:32
$begingroup$
The intuitive rule ( that must be used with some care) is always the same: '' use the Leibnitz notation as a fraction'' (also if it is not really a fraction). So we write: $frac{d}{dx}=frac{d}{dt}frac{dt}{dx}=frac{frac{d}{dt}}{frac{dx}{dt}}$
$endgroup$
– Emilio Novati
Jan 7 at 22:32
add a comment |
$begingroup$
I asked my teacher and figured it out: since we want to find the derivative of $frac{dy}{dx}$, we must find the derivative with respect to x, not t. Even though $frac{dy}{dx}$ is in terms of t, when we find the derivative it must be with respect to x. This should make sense. If you wanted to find the derivative of 15x, you would apply the $frac{d}{dx}$ operation, not something like $frac{d}{dt}$ or $frac{d}{dh}$ or any other variable. Since we now know we must use the $frac{d}{dx}$ on $frac{dy}{dx}$, it becomes clear that the $frac{frac{dy}{dt}}{frac{dx}{dt}}$ is the same as using $frac{d}{dx}$, since the dt's cancel.
answered Jan 7 at 22:22
AddisonAddison
314
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Intuitively (from Leibnitz notation):
$$
frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}=frac{d}{dt}left(frac{dy}{dx} right)frac{dt}{dx}
=
frac{d}{dx}left(frac{dy}{dx} right)
$$
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
$endgroup$
$begingroup$
Okay, so I see how the dt's cancel, creating the $frac{d}{dx}$. But why is the numerator so different than the denominator. When finding the first derivative, it is just $frac{frac{dy}{dt}}{frac{dx}{dt}}$, where the numerator and denominator are almost identical. Why is it that when you take the second derivative, the denominator looks so different from the numerator?
$endgroup$
– Addison
Jan 6 at 18:32
$begingroup$
The intuitive rule ( that must be used with some care) is always the same: '' use the Leibnitz notation as a fraction'' (also if it is not really a fraction). So we write: $frac{d}{dx}=frac{d}{dt}frac{dt}{dx}=frac{frac{d}{dt}}{frac{dx}{dt}}$
$endgroup$
– Emilio Novati
Jan 7 at 22:32
add a comment |
$begingroup$
Intuitively (from Leibnitz notation):
$$
frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}=frac{d}{dt}left(frac{dy}{dx} right)frac{dt}{dx}
=
frac{d}{dx}left(frac{dy}{dx} right)
$$
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
$endgroup$
$begingroup$
Okay, so I see how the dt's cancel, creating the $frac{d}{dx}$. But why is the numerator so different than the denominator. When finding the first derivative, it is just $frac{frac{dy}{dt}}{frac{dx}{dt}}$, where the numerator and denominator are almost identical. Why is it that when you take the second derivative, the denominator looks so different from the numerator?
$endgroup$
– Addison
Jan 6 at 18:32
$begingroup$
The intuitive rule ( that must be used with some care) is always the same: '' use the Leibnitz notation as a fraction'' (also if it is not really a fraction). So we write: $frac{d}{dx}=frac{d}{dt}frac{dt}{dx}=frac{frac{d}{dt}}{frac{dx}{dt}}$
$endgroup$
– Emilio Novati
Jan 7 at 22:32
add a comment |
$begingroup$
Intuitively (from Leibnitz notation):
$$
frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}=frac{d}{dt}left(frac{dy}{dx} right)frac{dt}{dx}
=
frac{d}{dx}left(frac{dy}{dx} right)
$$
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
$endgroup$
Intuitively (from Leibnitz notation):
$$
frac{frac{d}{dt}(frac{dy}{dx})}{frac{dx}{dt}}=frac{d}{dt}left(frac{dy}{dx} right)frac{dt}{dx}
=
frac{d}{dx}left(frac{dy}{dx} right)
$$
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
answered Jan 6 at 18:06
Emilio NovatiEmilio Novati
51.9k43474
51.9k43474
$begingroup$
Okay, so I see how the dt's cancel, creating the $frac{d}{dx}$. But why is the numerator so different than the denominator. When finding the first derivative, it is just $frac{frac{dy}{dt}}{frac{dx}{dt}}$, where the numerator and denominator are almost identical. Why is it that when you take the second derivative, the denominator looks so different from the numerator?
$endgroup$
– Addison
Jan 6 at 18:32
$begingroup$
The intuitive rule ( that must be used with some care) is always the same: '' use the Leibnitz notation as a fraction'' (also if it is not really a fraction). So we write: $frac{d}{dx}=frac{d}{dt}frac{dt}{dx}=frac{frac{d}{dt}}{frac{dx}{dt}}$
$endgroup$
– Emilio Novati
Jan 7 at 22:32
add a comment |
$begingroup$
Okay, so I see how the dt's cancel, creating the $frac{d}{dx}$. But why is the numerator so different than the denominator. When finding the first derivative, it is just $frac{frac{dy}{dt}}{frac{dx}{dt}}$, where the numerator and denominator are almost identical. Why is it that when you take the second derivative, the denominator looks so different from the numerator?
$endgroup$
– Addison
Jan 6 at 18:32
$begingroup$
The intuitive rule ( that must be used with some care) is always the same: '' use the Leibnitz notation as a fraction'' (also if it is not really a fraction). So we write: $frac{d}{dx}=frac{d}{dt}frac{dt}{dx}=frac{frac{d}{dt}}{frac{dx}{dt}}$
$endgroup$
– Emilio Novati
Jan 7 at 22:32
$begingroup$
Okay, so I see how the dt's cancel, creating the $frac{d}{dx}$. But why is the numerator so different than the denominator. When finding the first derivative, it is just $frac{frac{dy}{dt}}{frac{dx}{dt}}$, where the numerator and denominator are almost identical. Why is it that when you take the second derivative, the denominator looks so different from the numerator?
$endgroup$
– Addison
Jan 6 at 18:32
$begingroup$
Okay, so I see how the dt's cancel, creating the $frac{d}{dx}$. But why is the numerator so different than the denominator. When finding the first derivative, it is just $frac{frac{dy}{dt}}{frac{dx}{dt}}$, where the numerator and denominator are almost identical. Why is it that when you take the second derivative, the denominator looks so different from the numerator?
$endgroup$
– Addison
Jan 6 at 18:32
$begingroup$
The intuitive rule ( that must be used with some care) is always the same: '' use the Leibnitz notation as a fraction'' (also if it is not really a fraction). So we write: $frac{d}{dx}=frac{d}{dt}frac{dt}{dx}=frac{frac{d}{dt}}{frac{dx}{dt}}$
$endgroup$
– Emilio Novati
Jan 7 at 22:32
$begingroup$
The intuitive rule ( that must be used with some care) is always the same: '' use the Leibnitz notation as a fraction'' (also if it is not really a fraction). So we write: $frac{d}{dx}=frac{d}{dt}frac{dt}{dx}=frac{frac{d}{dt}}{frac{dx}{dt}}$
$endgroup$
– Emilio Novati
Jan 7 at 22:32
add a comment |
$begingroup$
I asked my teacher and figured it out: since we want to find the derivative of $frac{dy}{dx}$, we must find the derivative with respect to x, not t. Even though $frac{dy}{dx}$ is in terms of t, when we find the derivative it must be with respect to x. This should make sense. If you wanted to find the derivative of 15x, you would apply the $frac{d}{dx}$ operation, not something like $frac{d}{dt}$ or $frac{d}{dh}$ or any other variable. Since we now know we must use the $frac{d}{dx}$ on $frac{dy}{dx}$, it becomes clear that the $frac{frac{dy}{dt}}{frac{dx}{dt}}$ is the same as using $frac{d}{dx}$, since the dt's cancel.
answered Jan 7 at 22:22
AddisonAddison
314
$endgroup$
add a comment |
$begingroup$
I asked my teacher and figured it out: since we want to find the derivative of $frac{dy}{dx}$, we must find the derivative with respect to x, not t. Even though $frac{dy}{dx}$ is in terms of t, when we find the derivative it must be with respect to x. This should make sense. If you wanted to find the derivative of 15x, you would apply the $frac{d}{dx}$ operation, not something like $frac{d}{dt}$ or $frac{d}{dh}$ or any other variable. Since we now know we must use the $frac{d}{dx}$ on $frac{dy}{dx}$, it becomes clear that the $frac{frac{dy}{dt}}{frac{dx}{dt}}$ is the same as using $frac{d}{dx}$, since the dt's cancel.
answered Jan 7 at 22:22
AddisonAddison
314
$endgroup$
add a comment |
$begingroup$
I asked my teacher and figured it out: since we want to find the derivative of $frac{dy}{dx}$, we must find the derivative with respect to x, not t. Even though $frac{dy}{dx}$ is in terms of t, when we find the derivative it must be with respect to x. This should make sense. If you wanted to find the derivative of 15x, you would apply the $frac{d}{dx}$ operation, not something like $frac{d}{dt}$ or $frac{d}{dh}$ or any other variable. Since we now know we must use the $frac{d}{dx}$ on $frac{dy}{dx}$, it becomes clear that the $frac{frac{dy}{dt}}{frac{dx}{dt}}$ is the same as using $frac{d}{dx}$, since the dt's cancel.
answered Jan 7 at 22:22
AddisonAddison
314
$endgroup$
I asked my teacher and figured it out: since we want to find the derivative of $frac{dy}{dx}$, we must find the derivative with respect to x, not t. Even though $frac{dy}{dx}$ is in terms of t, when we find the derivative it must be with respect to x. This should make sense. If you wanted to find the derivative of 15x, you would apply the $frac{d}{dx}$ operation, not something like $frac{d}{dt}$ or $frac{d}{dh}$ or any other variable. Since we now know we must use the $frac{d}{dx}$ on $frac{dy}{dx}$, it becomes clear that the $frac{frac{dy}{dt}}{frac{dx}{dt}}$ is the same as using $frac{d}{dx}$, since the dt's cancel.
answered Jan 7 at 22:22
AddisonAddison
314
answered Jan 7 at 22:22
AddisonAddison
314
answered Jan 7 at 22:22
AddisonAddison
314
answered Jan 7 at 22:22
AddisonAddison
314
314
add a comment |
add a comment |
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