Let z = g(x, y) and suppose that x(t) = t^2 + 3t + 2 and y(t) = e^t + sin(3t)

Find (dz/dt)|t=0

if (dg/dx)|(1,2) = 6

(dg/dy)|(1,2) = -2

(dg/dx)|(2,1) = -3

(dg/dy)|(2,1) = 8

(dg/dx)|(0,0) = 0

(dg/dy)|(0,0) = -4

Thank You.

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- Dec 11th 2008, 01:38 PMDavidWagnerPartial derivative chain rule
Let z = g(x, y) and suppose that x(t) = t^2 + 3t + 2 and y(t) = e^t + sin(3t)

Find (dz/dt)|t=0

if (dg/dx)|(1,2) = 6

(dg/dy)|(1,2) = -2

(dg/dx)|(2,1) = -3

(dg/dy)|(2,1) = 8

(dg/dx)|(0,0) = 0

(dg/dy)|(0,0) = -4

Thank You. - Dec 11th 2008, 02:02 PMTheEmptySet
By the Chain rule we get

$\displaystyle \frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} +\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial g}{\partial x}(2t+3)+\frac{\partial g}{\partial y}(e^t+3\cos(3t)) $

Now we need to do some calculations

$\displaystyle x(0)=2 $ and $\displaystyle y(0)=1$

so $\displaystyle \frac{\partial g}{\partial x}\bigg |_{(2,1)}=-3$

and

$\displaystyle \frac{\partial g}{\partial y} \bigg |_{(2,1)}=8$

so Finally we get

$\displaystyle \frac{\partial z}{\partial t}=(-3)(2(0)+3)+8(e^0+3\cos(0))=-9+32=23$