< User talk:Hayson1991
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Revision as of 03:35, 14 November 2008 edit Hayson1991 (talk | contribs )242 edits →Clock Problem ← Previous edit
Revision as of 04:20, 14 November 2008 edit undo Hayson1991 (talk | contribs )242 edits →Piston speed : new sectionNext edit →
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<math>x=4\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br />
<math>x=4\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br />
<math>y=4\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br />
<math>y=4\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br />
== Piston speed ==
<math>x_w = r\cos\left(\theta\right)</math><br /><br />
<math>y_w = r\sin\left(\theta\right)</math><br /><br />
<math>D_p = \sqrt{L^2-x_w^2}+y_w</math><br /><br />
<math>\frac{dD_p}{dt} = \frac{1}{2}\left(L^2-x_w^2\right)^{-\frac{1}{2}}\cdot\left(-2x_w\frac{dx_w}{dt}\right)+\frac{dy_w}{dt}</math><br /><br />
<math>\frac{dx_w}{dt} = -r\sin\left(\theta\right)\cdot\omega</math><br /><br />
<math>\frac{dy_w}{dt} = r\cos\left(\theta\right)\cdot\omega</math><br /><br />
Revision as of 04:20, 14 November 2008
x
=
tan
(
y
)
{\displaystyle x=\tan \left(y\right)}
1
=
sec
2
(
y
)
∗
d
y
d
x
{\displaystyle 1=\sec ^{2}\left(y\right)*{\frac {dy}{dx}}}
(Chain rule, derivative of tan=sec^2)
1
sec
2
(
y
)
=
d
y
d
x
{\displaystyle {\frac {1}{\sec ^{2}\left(y\right)}}={\frac {dy}{dx}}}
cos
2
(
y
)
=
d
y
d
x
{\displaystyle \cos ^{2}\left(y\right)={\frac {dy}{dx}}}
d
y
d
x
=
cos
2
(
y
)
{\displaystyle {\frac {dy}{dx}}=\cos ^{2}\left(y\right)}
9~
x
2
y
+
x
y
2
=
6
{\displaystyle x^{2}y+xy^{2}=6\,}
(
2
x
∗
y
+
x
2
∗
d
y
d
x
)
+
(
1
∗
y
2
+
x
∗
2
y
d
y
d
x
)
=
0
{\displaystyle \left(2x*y+x^{2}*{\frac {dy}{dx}}\right)+\left(1*y^{2}+x*2y{\frac {dy}{dx}}\right)=0}
2
x
y
+
x
2
d
y
d
x
+
y
2
+
2
x
y
d
y
d
x
=
0
{\displaystyle 2xy+x^{2}{\frac {dy}{dx}}+y^{2}+2xy{\frac {dy}{dx}}=0}
x
2
d
y
d
x
+
2
x
y
d
y
d
x
=
−
2
x
y
−
y
2
{\displaystyle x^{2}{\frac {dy}{dx}}+2xy{\frac {dy}{dx}}=-2xy-y^{2}}
d
y
d
x
=
−
2
x
y
−
y
2
x
2
+
2
x
y
{\displaystyle {\frac {dy}{dx}}={\frac {-2xy-y^{2}}{x^{2}+2xy}}}
d
y
d
x
=
−
2
x
y
+
y
2
x
2
+
2
x
y
{\displaystyle {\frac {dy}{dx}}=-{\frac {2xy+y^{2}}{x^{2}+2xy}}}
Multiple u's
To Find dy/dx for
y
=
2
cos
(
(
5
x
)
2
)
{\displaystyle y=2\cos \left(\left(5x\right)^{2}\right)}
The way she explains it
you'll make 3 u's
Let
u
=
2
cos
(
u
)
{\displaystyle {\text{Let }}u=2\cos \left(u\right)}
Let
u
=
u
2
{\displaystyle {\text{Let }}u=u^{2}\,}
Let
u
=
5
x
{\displaystyle {\text{Let }}u=5x\,}
Gaaah, help~~
Find
d
y
d
x
{\displaystyle {\frac {dy}{dx}}\,}
then find
d
2
y
d
x
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}\,}
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1\,}
2
x
+
2
y
d
y
d
x
=
0
{\displaystyle 2x+2y{\frac {dy}{dx}}=0\,}
Find first derivative
d
y
d
x
=
−
2
x
2
y
{\displaystyle {\frac {dy}{dx}}={\frac {-2x}{2y}}\,}
d
y
d
x
=
−
x
y
{\displaystyle {\frac {dy}{dx}}=-{\frac {x}{y}}\,}
Find second derivative
2
+
(
2
d
y
d
x
∗
d
y
d
x
+
2
y
∗
d
2
y
d
x
2
)
=
0
{\displaystyle 2+\left(2{\frac {dy}{dx}}*{\frac {dy}{dx}}+2y*{\frac {d^{2}y}{dx^{2}}}\right)=0\,}
2
(
d
y
d
x
)
2
+
2
y
d
2
y
d
x
2
=
−
2
{\displaystyle 2\left({\frac {dy}{dx}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,}
2
(
−
x
y
)
2
+
2
y
d
2
y
d
x
2
=
−
2
{\displaystyle 2\left(-{\frac {x}{y}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,}
2
x
2
y
2
+
2
y
d
2
y
d
x
2
=
−
2
{\displaystyle 2{\frac {x^{2}}{y^{2}}}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,}
2
y
d
2
y
d
x
2
=
−
2
−
2
x
2
y
2
{\displaystyle 2y{\frac {d^{2}y}{dx^{2}}}=-2-2{\frac {x^{2}}{y^{2}}}\,}
d
2
y
d
x
2
=
−
2
−
2
x
2
y
2
2
y
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {-2-2{\frac {x^{2}}{y^{2}}}}{2y}}\,}
d
2
y
d
x
2
=
−
1
y
−
x
2
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y}}-{\frac {x^{2}}{y^{3}}}\,}
d
2
y
d
x
2
=
−
y
2
y
3
−
x
2
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}}{y^{3}}}-{\frac {x^{2}}{y^{3}}}\,}
d
2
y
d
x
2
=
−
y
2
+
x
2
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}+x^{2}}{y^{3}}}\,}
d
2
y
d
x
2
=
−
1
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y^{3}}}\,}
Clock Problem ~
minute hand
x
=
5
cos
(
π
2
−
t
⋅
2
π
60
)
{\displaystyle x=5\cos \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{60}}\right)}
y
=
5
sin
(
π
2
−
t
⋅
2
π
60
)
{\displaystyle y=5\sin \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{60}}\right)}
hour hand
x
=
4
cos
(
π
2
−
t
⋅
2
π
12
)
{\displaystyle x=4\cos \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{12}}\right)}
y
=
4
sin
(
π
2
−
t
⋅
2
π
12
)
{\displaystyle y=4\sin \left({\frac {\pi }{2}}-{\frac {t\cdot 2\pi }{12}}\right)}
Piston speed
x
w
=
r
cos
(
θ
)
{\displaystyle x_{w}=r\cos \left(\theta \right)}
y
w
=
r
sin
(
θ
)
{\displaystyle y_{w}=r\sin \left(\theta \right)}
D
p
=
L
2
−
x
w
2
+
y
w
{\displaystyle D_{p}={\sqrt {L^{2}-x_{w}^{2}}}+y_{w}}
d
D
p
d
t
=
1
2
(
L
2
−
x
w
2
)
−
1
2
⋅
(
−
2
x
w
d
x
w
d
t
)
+
d
y
w
d
t
{\displaystyle {\frac {dD_{p}}{dt}}={\frac {1}{2}}\left(L^{2}-x_{w}^{2}\right)^{-{\frac {1}{2}}}\cdot \left(-2x_{w}{\frac {dx_{w}}{dt}}\right)+{\frac {dy_{w}}{dt}}}
d
x
w
d
t
=
−
r
sin
(
θ
)
⋅
ω
{\displaystyle {\frac {dx_{w}}{dt}}=-r\sin \left(\theta \right)\cdot \omega }
d
y
w
d
t
=
r
cos
(
θ
)
⋅
ω
{\displaystyle {\frac {dy_{w}}{dt}}=r\cos \left(\theta \right)\cdot \omega }