Mixing
Differential Equation for a curve from a given point.
$begingroup$
Question,
The y intercept of the tangent line to a curve at any point is always
equal to the slope at that point. If the curve passes through the point
(2, 1), find the equation of the curve.
Based off this little information I have $y=y'x+b$ where $b$ would be $y'$ evaluated at the point. I'm not sure how to get this into an equation for a curve or if there is a much better approach. Thanks in advance.
ordinary-differential-equations
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
$endgroup$
add a comment |
$begingroup$
Question,
The y intercept of the tangent line to a curve at any point is always
equal to the slope at that point. If the curve passes through the point
(2, 1), find the equation of the curve.
Based off this little information I have $y=y'x+b$ where $b$ would be $y'$ evaluated at the point. I'm not sure how to get this into an equation for a curve or if there is a much better approach. Thanks in advance.
ordinary-differential-equations
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
$endgroup$
add a comment |
$begingroup$
Question,
The y intercept of the tangent line to a curve at any point is always
equal to the slope at that point. If the curve passes through the point
(2, 1), find the equation of the curve.
Based off this little information I have $y=y'x+b$ where $b$ would be $y'$ evaluated at the point. I'm not sure how to get this into an equation for a curve or if there is a much better approach. Thanks in advance.
ordinary-differential-equations
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
$endgroup$
Question,
The y intercept of the tangent line to a curve at any point is always
equal to the slope at that point. If the curve passes through the point
(2, 1), find the equation of the curve.
Based off this little information I have $y=y'x+b$ where $b$ would be $y'$ evaluated at the point. I'm not sure how to get this into an equation for a curve or if there is a much better approach. Thanks in advance.
ordinary-differential-equations
ordinary-differential-equations
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
asked Dec 11 '13 at 7:42
slow_Mathslow_Math
64
64
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you fix a point $x_0$ the equation of the tangent line in $(x_0,f(x_0))$ is
$$
y = f(x_0) + f'(x_0)(x-x_0) = f'(x_0) x + f(x_0) - f'(x_0) x_0
$$
hence $q=f(x_0)-f'(x_0)x_0$ is the intercept, and the equation is (call $x=x_0$, $y=f(x_0)$
$$
y' = y - y'x
$$
i.e.
$$
y'(1+x) = y
$$
i.e.
$$
frac{y'}{y} = frac{1}{1+x}
$$
Can you solve this?
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
$endgroup$
$begingroup$
Thanks! Yeah I believe so, trying it now, integrating both sides and solving for y.
$endgroup$
– slow_Math
Dec 11 '13 at 8:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you fix a point $x_0$ the equation of the tangent line in $(x_0,f(x_0))$ is
$$
y = f(x_0) + f'(x_0)(x-x_0) = f'(x_0) x + f(x_0) - f'(x_0) x_0
$$
hence $q=f(x_0)-f'(x_0)x_0$ is the intercept, and the equation is (call $x=x_0$, $y=f(x_0)$
$$
y' = y - y'x
$$
i.e.
$$
y'(1+x) = y
$$
i.e.
$$
frac{y'}{y} = frac{1}{1+x}
$$
Can you solve this?
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
$endgroup$
$begingroup$
Thanks! Yeah I believe so, trying it now, integrating both sides and solving for y.
$endgroup$
– slow_Math
Dec 11 '13 at 8:01
add a comment |
$begingroup$
If you fix a point $x_0$ the equation of the tangent line in $(x_0,f(x_0))$ is
$$
y = f(x_0) + f'(x_0)(x-x_0) = f'(x_0) x + f(x_0) - f'(x_0) x_0
$$
hence $q=f(x_0)-f'(x_0)x_0$ is the intercept, and the equation is (call $x=x_0$, $y=f(x_0)$
$$
y' = y - y'x
$$
i.e.
$$
y'(1+x) = y
$$
i.e.
$$
frac{y'}{y} = frac{1}{1+x}
$$
Can you solve this?
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
$endgroup$
$begingroup$
Thanks! Yeah I believe so, trying it now, integrating both sides and solving for y.
$endgroup$
– slow_Math
Dec 11 '13 at 8:01
add a comment |
$begingroup$
If you fix a point $x_0$ the equation of the tangent line in $(x_0,f(x_0))$ is
$$
y = f(x_0) + f'(x_0)(x-x_0) = f'(x_0) x + f(x_0) - f'(x_0) x_0
$$
hence $q=f(x_0)-f'(x_0)x_0$ is the intercept, and the equation is (call $x=x_0$, $y=f(x_0)$
$$
y' = y - y'x
$$
i.e.
$$
y'(1+x) = y
$$
i.e.
$$
frac{y'}{y} = frac{1}{1+x}
$$
Can you solve this?
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
$endgroup$
If you fix a point $x_0$ the equation of the tangent line in $(x_0,f(x_0))$ is
$$
y = f(x_0) + f'(x_0)(x-x_0) = f'(x_0) x + f(x_0) - f'(x_0) x_0
$$
hence $q=f(x_0)-f'(x_0)x_0$ is the intercept, and the equation is (call $x=x_0$, $y=f(x_0)$
$$
y' = y - y'x
$$
i.e.
$$
y'(1+x) = y
$$
i.e.
$$
frac{y'}{y} = frac{1}{1+x}
$$
Can you solve this?
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
answered Dec 11 '13 at 7:48
Emanuele PaoliniEmanuele Paolini
17.9k22052
17.9k22052
$begingroup$
Thanks! Yeah I believe so, trying it now, integrating both sides and solving for y.
$endgroup$
– slow_Math
Dec 11 '13 at 8:01
add a comment |
$begingroup$
Thanks! Yeah I believe so, trying it now, integrating both sides and solving for y.
$endgroup$
– slow_Math
Dec 11 '13 at 8:01
$begingroup$
Thanks! Yeah I believe so, trying it now, integrating both sides and solving for y.
$endgroup$
– slow_Math
Dec 11 '13 at 8:01
$begingroup$
Thanks! Yeah I believe so, trying it now, integrating both sides and solving for y.
$endgroup$
– slow_Math
Dec 11 '13 at 8:01
add a comment |
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