Find the equation of the curve $y''=2x+1$
up vote
0
down vote
favorite
I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
asked Nov 18 at 8:32
Holly Millican
528
add a comment |
up vote
0
down vote
favorite
I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
asked Nov 18 at 8:32
Holly Millican
528
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
asked Nov 18 at 8:32
Holly Millican
528
I am stuck on the last step of the following equation:
If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.
So far I have the following:
If $y''=2x+1$
$y'=x^2+x+c$
then use the given point to solve for $c$ as follows:
$0=3^2+3+c$
$c=-12$
so, if $y'=x^2+x-12$
$y= frac{x^3}{3}+frac{x^2}{2}-12x+c$
now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!
differential-equations curves
differential-equations curves
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
asked Nov 18 at 8:32
Holly Millican
528
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
asked Nov 18 at 8:32
Holly Millican
528
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
edited Nov 18 at 10:56
Kelvin Lois
3,0772822
3,0772822
asked Nov 18 at 8:32
Holly Millican
528
asked Nov 18 at 8:32
Holly Millican
528
asked Nov 18 at 8:32
Holly Millican
528
528
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
add a comment |
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
It should pass through the given point
– Semsem
Nov 18 at 8:43
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,635815
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,635815
add a comment |
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,635815
add a comment |
up vote
1
down vote
up vote
1
down vote
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,635815
It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain begin{align}y'(x)&=x^2+x+c\y(x)&=frac{1}{3}x^3+frac{1}{2}x^2+cx+dend{align}
From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $frac{49}{2}$.
answered Nov 18 at 9:06
Fakemistake
1,635815
answered Nov 18 at 9:06
Fakemistake
1,635815
answered Nov 18 at 9:06
Fakemistake
1,635815
answered Nov 18 at 9:06
Fakemistake
1,635815
1,635815
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003264%2ffind-the-equation-of-the-curve-y-2x1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
It should pass through the given point
– Semsem
Nov 18 at 8:43
but the given point is a stationary point, so wouldn't it only apply to the first derivative?
– Holly Millican
Nov 18 at 8:45
1
Use $y(3)=2$ for determining the constant
– Fakemistake
Nov 18 at 8:46
thank you, that got me where I needed
– Holly Millican
Nov 18 at 8:52