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Questions and Answers
Is the equation $u = f(2x - 3y, 3y - 4z, 4z - 2x)$ equivalent to $2 \frac{\partial u},{\partial x} + 3 \frac{\partial u},{\partial y} + 2 \frac{\partial u},{\partial z} = c \frac{\partial (x,y)},{\partial (x,y,z)}$?
Is the equation $u = f(2x - 3y, 3y - 4z, 4z - 2x)$ equivalent to $2 \frac{\partial u},{\partial x} + 3 \frac{\partial u},{\partial y} + 2 \frac{\partial u},{\partial z} = c \frac{\partial (x,y)},{\partial (x,y,z)}$?
False
If $x = r \cos\theta$, $y = \sin\theta$, then is $\frac{\partial (r,\theta)},{\partial (x,y)} = 1$?
If $x = r \cos\theta$, $y = \sin\theta$, then is $\frac{\partial (r,\theta)},{\partial (x,y)} = 1$?
True
If $x = u(1+v)$, $y = v(1+u)$, then is $\frac{\partial (u,v)},{\partial (u,v)} = 1$?
If $x = u(1+v)$, $y = v(1+u)$, then is $\frac{\partial (u,v)},{\partial (u,v)} = 1$?
True
If $u = 2xy$, $v = x^2 - y^2$, $x = r\cos\theta$, $y = r\sin\theta$, then is $\frac{\partial (u,v)},{\partial (r,\theta)} = 2r$?
If $u = 2xy$, $v = x^2 - y^2$, $x = r\cos\theta$, $y = r\sin\theta$, then is $\frac{\partial (u,v)},{\partial (r,\theta)} = 2r$?
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Is the maximum value of $x+y+z$ subject to the condition $bc + y + z = 1$ equal to 1?
Is the maximum value of $x+y+z$ subject to the condition $bc + y + z = 1$ equal to 1?
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Find the Taylor series of $f(x, y) = 1 - x - y$ around $(0,0)$.
Find the Taylor series of $f(x, y) = 1 - x - y$ around $(0,0)$.
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Expand $f(x, y) = e^x \sin y$ in power of $x$ and $y$ up to the third degree.
Expand $f(x, y) = e^x \sin y$ in power of $x$ and $y$ up to the third degree.
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Find the maximum value of $x + y + z$ subject to the condition $bc + y + z = 1$.
Find the maximum value of $x + y + z$ subject to the condition $bc + y + z = 1$.
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Discuss the maxima and minima value of the function $f(x, y) = x^3 + y^3 - 12x - 3y + 20$.
Discuss the maxima and minima value of the function $f(x, y) = x^3 + y^3 - 12x - 3y + 20$.
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If $z = f(x, y)$, $x = u\cos\alpha - v\sin\alpha$, $y = u\cos\alpha + v\cos\alpha$, prove that $\frac{\partial z},{\partial u} = \frac{\partial z},{\partial v} = \cos\alpha - \sin\alpha$.
If $z = f(x, y)$, $x = u\cos\alpha - v\sin\alpha$, $y = u\cos\alpha + v\cos\alpha$, prove that $\frac{\partial z},{\partial u} = \frac{\partial z},{\partial v} = \cos\alpha - \sin\alpha$.
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