Tutorial 3: Partial Derivatives & Engineering Applications of Partial Derivatives
Q1: Find the partial derivatives \((\frac{\partial f}{\partial y})\) and \((\frac{\partial f}{\partial x})\) of these functions using the limit definition.
(a) \(f(x,y)=x^2-4xy+y^2\) (b) \(f(x,y)=2x^3+3xy-y^2\)
Solution
Question (a)
\[\begin{aligned} \frac{\partial f}{\partial x}&= \lim_{h \rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h} \\ &= \lim_{h \rightarrow 0} \frac{(x+h)^2-4(x+h)y+y^2-(x^2-4xy+y^2)}{h} \\ &= \lim_{h \rightarrow 0} \frac{x^2+2xh+h^2-4xy-4hy+y^2-(x^2-4xy+y^2)}{h} \\ &= \lim_{h \rightarrow 0} \frac{2xh+h^2-4hy}{h} \\ &= \lim_{h \rightarrow 0} (2x+h-4y) \\ &= \boldsymbol{2x-4y} \end{aligned}\] \[\begin{aligned} \frac{\partial f}{\partial y}&= \lim_{h \rightarrow 0} \frac{f(x,y+h)-f(x,y)}{h} \\ &= \lim_{h \rightarrow 0} \frac{x^2-4x(y+h)+(y+h)^2-(x^2-4xy+y^2)}{h} \\ &= \lim_{h \rightarrow 0} \frac{x^2-4xy-4xh+y^2+2yh+h^2-(x^2-4xy+y^2)}{h} \\ &= \lim_{h \rightarrow 0} \frac{-4xh+2yh+h^2}{h} \\ &= \lim_{h \rightarrow 0} (-4x+2y+h) \\ &= \boldsymbol{-4x+2y} \end{aligned}\]Question (b)
\[\begin{aligned} \frac{\partial f}{\partial x} &= \lim_{h \rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h} \\ &= \lim_{h \rightarrow 0} \frac{h(6x^2+6xh+2h^2+3y)}{h} \\ &= \lim_{h \rightarrow 0} (6x^2+6xh+2h^2+3y) \\ &= \boldsymbol{6x^2+3y} \end{aligned}\] \[\begin{aligned} \frac{\partial f}{\partial y} &= \lim_{h \rightarrow 0} \frac{f(x,y+h)-f(x,y)}{h} \\ &= \lim_{h \rightarrow 0} \frac{h(3x-2y-h)}{h} \\ &= \lim_{h \rightarrow 0} (3x-2y-h) \\ &= \boldsymbol{3x-2y} \end{aligned}\]Q2: Determine all the first and second order partial derivatives of the function
(a) \(f(x,y) = x^2y^3+3y+x\)
(b) \(f(x,y) = y\sin{x} + x\cos{y}\)
(c) \(f(x,y) = x^4 \sin {3y}\)
(d) \(f(x,y) = e^{xy} (2x - y)\)
(e) \(f(x,y,z) = z^2e^{xy} + x\cos{(y^2z)}\)
Solution
Question (a)
\[\begin{aligned} \frac{\partial f}{\partial x}&= \boldsymbol{2xy^3+1} \\ \frac{\partial f}{\partial y}&= \boldsymbol{3x^2y^2+3} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{2y^3} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{6xy^2} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{6xy^2} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{6x^2y} \\ \end{aligned}\]Question (b)
\[\begin{aligned} \frac{\partial f}{\partial x}&= \boldsymbol{y\cos{x}+\cos{y}} \\ \frac{\partial f}{\partial y}&= \boldsymbol{\sin{x}-x\sin{y}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{-y\sin{x}} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{\cos{x}-\sin{y}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{\cos{x}-\sin{y}} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{-x\cos{y}} \\ \end{aligned}\]Question (c)
\[\begin{aligned} \frac{\partial f}{\partial x}&= \boldsymbol{4x^3\sin{3y}} \\ \frac{\partial f}{\partial y}&= \boldsymbol{3x^4\cos{3y}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{12x^2\sin{3y}} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{12x^3\cos{3y}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{12x^3\cos{3y}} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{-9x^4\sin{3y}} \\ \end{aligned}\]Question (d)
\[\begin{aligned} \frac{\partial f}{\partial x}&= \boldsymbol{e^{xy}(2xy-y^2+2)} \\ \frac{\partial f}{\partial y}&= \boldsymbol{e^{xy}(2x^2-xy-1)} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{e^{xy}(2xy^2-y^3+4y)} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{e^{xy}(2x^3y-xy^2+4x-2y)} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{e^{xy}(2x^2y-xy^2+4x-2y)} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{e^{xy}(2x^3-x^2y-2x)} \\ \end{aligned}\]Question (e)
\[\begin{aligned} \frac{\partial f}{\partial x}&= \boldsymbol{yz^2e^{xy}+\cos{(y^2z)}} \\ \frac{\partial f}{\partial y}&= \boldsymbol{xz^2e^{xy}-2xyz\sin{(y^2z)}} \\ \frac{\partial f}{\partial z}&= \boldsymbol{2ze^{xy}-xy^2\sin{(y^2z)}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{y^2z^2e^{xy}} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{x^2z^2e^{xy}-2xz\sin{(y^2z)-4xy^2z^2\cos{(y^2z)}}} \\ \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial z}\right)&= \boldsymbol{2e^{xy}-xy^4\cos{(y^2z)}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)&=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)= \boldsymbol{z^{2} e^{xy} +z^{2} xye^{xy} -2yz\sin\left( y^{2} z\right)} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial z}\right)&= \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial y}\right) = \boldsymbol{2xze^{xy}-2xy\sin{(y^2z)}-2xy^3z\cos{(y^2z)}} \\ \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial x}\right)&= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial z}\right) = \boldsymbol{2yze^{xy}-y^2\sin{(y^2z)}} \\ \end{aligned}\]Q3: Find both partial derivatives for each of the following two variables functions.
(a) \(f(x,y) = \log{x} + 3y + 1\)
(b) \(f(x,y) = ye^{x+y}\)
(c) \(h(x,y) = x\sin{y}-y\cos{x}\)
(d) \(p(x,y) = x^y+y^2\)
(e) \(U(x,y) = \frac{9y^3}{x-y}\)
Solution
Question (a)
\[\begin{aligned} \frac{df}{dx}&=\boldsymbol{\frac{1}{x}} \\ \frac{df}{dy}&=\boldsymbol{3} \end{aligned}\]Question (b)
\[\begin{aligned} \frac{dg}{dx}&=\boldsymbol{ye^{x+y}} \\ \frac{dg}{dy}&=\boldsymbol{e^{x+y}+ye^{x+y}} \end{aligned}\]Question (c)
\[\begin{aligned} \frac{dh}{dx}&=\boldsymbol{\sin{y}+y\sin{x}} \\ \frac{dh}{dy}&=\boldsymbol{x\cos{y}-\cos{x}} \end{aligned}\]Question (d)
\[\begin{aligned} \frac{dp}{dx}&=\boldsymbol{yx^{y-1}} \\ \frac{dp}{dy}&=\boldsymbol{x^y\ln{x}+2y} \end{aligned}\]Question (e)
\[\begin{aligned} \frac{dU}{dx}&=\frac{(x-y)(0)-9y^3(1)}{(x-y)^2}=\boldsymbol{\frac{-9y^3}{(x-y)^2}} \\ \frac{dU}{dy}&=\frac{(x-y)(27y^2)-9y^3(-1)}{(x-y)^2}=\boldsymbol{\frac{27xy^2-18y^3}{(x-y)^2}} \end{aligned}\]Q4: Compute the first and second partial derivatives of \(z\).
\[z=\frac{x}{y^2}+\frac{1}{x^3}+\log{(x+y)}\]Solution
\[\begin{aligned} \frac{\partial z}{\partial x}&=\boldsymbol{\frac{1}{y^2}-\frac{3}{x^4}+\frac{1}{x+y}} \\ \frac{\partial z}{\partial y}&=\boldsymbol{\frac{-2x}{y^3}+\frac{1}{x+y}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{\frac{12}{x^5}+\frac{-1}{(x+y)^2}} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)&= \boldsymbol{\frac{-2}{y^3}+\frac{-1}{(x+y)^2}} \\ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{\frac{-2}{y^3}+\frac{-1}{(x+y)^2}} \\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)&= \boldsymbol{\frac{6x}{y^4}+\frac{-1}{(x+y)^2}} \\ \end{aligned}\]Q5: For \(f(x,y,z)\), use the implicit function theorem to find \(\frac{dy}{dx}\) and \(\frac{dy}{dz}\):
(a) \(f(x,y,z) = x^2y^3+z^2+xyz\)
(b) \(f(x,y,z) = x^3z^2+y^3+4xyz\)
(c) \(f(x,y,z) = 3x^2y^3+xz^2y^2+y^3zx^4+y^2z\)
Solution
Question (a)
\[\begin{aligned} \frac{dy}{dx}&=-\frac{fx}{fy}=\boldsymbol{-\frac{2xy^3+yz}{3x^2y^2+xz}} \\ \frac{dy}{dz}&=-\frac{fz}{fy}=\boldsymbol{-\frac{2z+xy}{3x^2y^2+xz}} \\ \end{aligned}\]Question (b)
\[\begin{aligned} \frac{dy}{dx}&=-\frac{fx}{fy}=\boldsymbol{-\frac{3x^2z^2+4yz}{3y^2+4xz}} \\ \frac{dy}{dz}&=-\frac{fz}{fy}=\boldsymbol{-\frac{2x^3z+4xy}{3y^2+4xz}} \\ \end{aligned}\]Question (c)
\[\begin{aligned} \frac{dy}{dx}&=-\frac{fx}{fy}=\boldsymbol{-\frac{6xy^3+y^2z^2+4x^3y^3z}{9x^2y^2+2xyz^2+3x^4y^2z+2yz}} \\ \frac{dy}{dz}&=-\frac{fz}{fy}=\boldsymbol{-\frac{2xy^2z+x^4y^3+y^2}{9x^2y^2+2xyz^2+3x^4y^2z+2yz}} \\ \end{aligned}\]Q6: Find \(\frac{\partial F}{\partial s}\) and \(\frac{\partial F}{\partial t}\),if applicable, for the following composite functions.
(a) \(F = \sin (x + y)\) where \(x = 2st\) and \(y = s^2 + t^2\).
(b) \(F = e^x\cos y\) where \(x = s^2 - t^2\) and \(y = 2st\).
(c) \(F = 5x - 3y^2 + 7z^3\) where \(x = 2s + 3t\), \(y = s - t\) and \(z = 4s + t\).
(d) \(F = \ln (x^2 + y)\) where \(x = \exp (s+t^2)\) and \(y = s^2 + t\).
(e) \(F = x^2y^2\) where \(x = s \cos t\) and \(y = s \sin t\).
(f) \(F = xy + yz^2\) where \(x = e^t\), \(y = e^t \sin t\) and \(z = e^t \cos t\).
Solution:
Question (a)
\[\begin{aligned} \frac{\partial F}{\partial s}&=\frac{\partial F}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial F}{\partial y} \frac{\partial y}{\partial s} \\ &=\cos (x+y)(2t)+\cos (x+y)(2s) \\ &=2t\cos (2st+s^2+t^2)+2s\cos (st+s^2+t^2) \\ &= \boldsymbol{2\cos ((s+t)^2)(s+t)} \end{aligned}\] \[\begin{aligned} \frac{\partial F}{\partial t}&=\frac{\partial F}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial F}{\partial y} \frac{\partial y}{\partial t} \\ &=\cos (x+y)(2s)+\cos (x+y)(2t) \\ &= \boldsymbol{2\cos ((s+t)^2)(s+t)} \end{aligned}\]Question (b)
\[\begin{aligned} \frac{\partial F}{\partial s}&=\frac{\partial F}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial F}{\partial y} \frac{\partial y}{\partial s} \\ &=e^x\cos{y}(2s)+(-\sin y)e^x(2t) \\ &= 2se^x\cos y-2te^x\sin y \\ &= \boldsymbol{2se^{s^2-t^2}\cos 2st-2te^{s^2-t^2}\sin 2st} \end{aligned}\] \[\begin{aligned} \frac{\partial F}{\partial t}&=\frac{\partial F}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial F}{\partial y} \frac{\partial y}{\partial t} \\ &=e^x\cos{y}(2t)+(-\sin y)e^x(2s) \\ &= 2te^x\cos y-2se^x\sin y \\ &= \boldsymbol{-2te^{s^2-t^2}\cos 2st-2se^{s^2-t^2}\sin 2st} \end{aligned}\]Question (c)
\[\begin{aligned} \frac{\partial F}{\partial s}&=\frac{\partial F}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial F}{\partial y} \frac{\partial y}{\partial s}+\frac{\partial F}{\partial z} \frac{\partial z}{\partial s} \\ &=5(2)+(-6y)(1)+21z^2(4)\\ &=10-6y+84z^2\\ &=10-6(s-t)+84(4s+t)^2\\ &=10-6s+6t+84(16s^2+8st+t^2)\\ &= \boldsymbol{10-6s-6t+1344s^2+672st+84t^2} \end{aligned}\] \[\begin{aligned} \frac{\partial F}{\partial t}&=\frac{\partial F}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial F}{\partial y} \frac{\partial y}{\partial t}+\frac{\partial F}{\partial z} \frac{\partial z}{\partial t} \\ &=5(3)-6y(-1)+21z^2(1) \\ &=15+6y+21z^2\\ &=15+6(s-t)+21(4s-t)^2\\ &=15+6s-6t+21(16s^2+8st+t^2)\\ &= \boldsymbol{15+6s-6t+336s^2+168st+21t^2} \end{aligned}\]Question (d)
\[\begin{aligned} \frac{\partial F}{\partial s} & =\frac{\partial F}{\partial x}\frac{\partial x}{\partial s} +\frac{\partial F}{\partial y}\frac{\partial y}{\partial s}\\ & =\frac{2x}{x^{2} +y} e^{s+t^{2}} +\frac{1}{x^{2} +y} 2s\\ & =\mathbf{\frac{2}{x^{2} +y}\left( xe^{s+t^{2}} +s\right)} \end{aligned}\] \[\begin{aligned} \frac{\partial F}{\partial t} & =\frac{\partial F}{\partial x}\frac{\partial x}{\partial t} +\frac{\partial F}{\partial y}\frac{\partial y}{\partial t}\\ & =\frac{2x}{x^{2} +y} 2te^{s+t^{2}} +\frac{1}{x^{2} +y}( 1)\\ & =\mathbf{\frac{1}{x^{2} +y}\left( 4xte^{s+t^{2}} +1\right)} \end{aligned}\]Question (e)
\[\begin{aligned} \frac{\partial F}{\partial s} & =\frac{\partial F}{\partial x}\frac{\partial x}{\partial s} +\frac{\partial F}{\partial y}\frac{\partial y}{\partial s}\\ & =2xy^{2}\cos t+2x^{2} y\sin t\\ & =2( s\cos t)\left( s^{2}\sin^{2} t\right)(\cos t) +2\left( s^{2}\cos^{2} t\right)( s\sin t)(\sin t)\\ & =2s^{3}\cos^{2} t\sin^{2} t+2s^{3}\cos^{2} t\sin^{2} t\\ & =4s^{3}\cos^{2} t\sin^{2} t\\ & =4s^{3}(\cos t\sin t)^{2}\\ & =4s^{3}\left(\frac{1}{2}\sin 2t\right)^{2}\\ & =\mathbf{s^{3}(\sin 2t)^{2}} \end{aligned}\] \[\begin{aligned} \frac{\partial F}{\partial t} & =\frac{\partial F}{\partial x}\frac{\partial x}{\partial t} +\frac{\partial F}{\partial y}\frac{\partial y}{\partial t}\\ & =2xy^{2}( -s\sin t) +2x^{2} y( s\cos t)\\ & =2x^{2} ys\cos t-2xy^{2} s\sin t\\ & =2\left( s^{2}\cos^{2} t\right)( s\sin t) s\cos t-2( s\cos t)\left( s^{2}\sin^{2} t\right) s\sin t\\ & =2s^{4}\cos^{3} t\sin t-2s^{4}\cos t\sin^{3} t\\ & =2s^{4}(\cos t\sin t)\left(\cos^{2} t-\sin^{2} t\right)\\ & =2s^{4}\left(\frac{1}{2}\sin 2t\right)(\cos 2t)\\ & =\mathbf{s^{4}\sin 2t\cos 2t} \end{aligned}\]Question (f)
\[\begin{aligned} \frac{dF}{dt} & =\frac{\partial F}{\partial x}\frac{\partial x}{\partial t} +\frac{\partial F}{\partial y}\frac{\partial y}{\partial t} +\frac{\partial F}{\partial z}\frac{\partial z}{\partial t}\\ & =ye^{t} +\left( x+z^{2}\right)\left( e^{t}\cos t+\sin^{t} e^{t}\right) +2yz\left( -e^{t}\sin t+\cos te^{t}\right)\\ & =ye^{t} +xe^{t}\cos t+x\sin^{t} e^{t} +z^{2} e^{t}\cos t+z^{2}\sin te^{t} -2yze^{t}\sin t+2yz\cos te^{t}\\ & =ye^{t} +e^{t}\cos t\left( x+z^{2} +2yz\right) +e^{t}\sin t\left( x+z^{2} -2yz\right)\\ & =e^{t}\sin t\cdot e^{t} +e^{t}\cos t\left( e^{t} +e^{2t}\cos t+2e^{2t}\sin t\cos t\right)\\ & \ \ \ \ \ +e^{t}\sin t\left( e^{t} +e^{2t}\cos^{2} t-2e^{2t}\sin t\cos t\right)\\ & =e^{2t}\sin t+e^{2t}\cos t+e^{3t}\cos^{3} t+2e^{2t}\sin t\cos^{2} t+e^{2t}\sin t+e^{3t}\sin t\cos^{2} t\\ & \ \ \ \ \ -2e^{3} t\sin^{2} t\cos t\\ & =2e^{2t}\sin t+e^{2t}\cos t+e^{3t}\cos^{3} t+3e^{3t}\sin t\cos^{2} t-2e^{3t}\sin^{2} t\cos t\\ & =\mathbf{e^{2t}( 2\sin t+\cos t) +e^{3t}\left(\cos^{3} t+3\sin t\cos^{2} t-2\sin^{2} t\cos t\right)} \end{aligned}\]Q7: Find \(\frac{dy}{dx}\) and \(\frac{dy}{dz}\) (if applicable) for each of the following.
(a) \(y^{x} +1=0\)
(b) \(7x^{2} +2xy^{2} +9y^{4} =0\)
(c) \(x^{3} z^{2} +y^{3} +4xyz=0\)
(d) \(3x^{2} y^{3} +xz^{2} y^{2} +y^{3} zx^{4} +y^{2} z=0\)
(e) \(y^{5} +x^{2} y^{3} =1+y\exp\left( x^{2}\right)\)
Solution
Question (a)
\[\begin{aligned} \frac{dy}{dx} & =-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} =-\frac{-y^{x}\ln y}{xy^{x-1}} =\mathbf{-\frac{y}{x}\ln y} \end{aligned}\]Let \(z=y^{x} \Longrightarrow \ln z=x\ln y\),
\[\begin{aligned} \frac{1}{z}\frac{dy}{dx} & =\ln y\\ \frac{dy}{dx} & =z\ln y\\ & =\mathbf{y^{2}\ln y} \end{aligned}\]Question (b)
\[\begin{aligned} \frac{dy}{dx} & =-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} =\mathbf{-\frac{14x+2y^{2}}{36y^{2} +4xy}} \end{aligned}\]Question (c)
\[\begin{aligned} \frac{dy}{dx} & =-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} =\mathbf{-\frac{3x^{2} z^{2} +4yz}{3y^{2} +4xz}}\\ \frac{dy}{dz} & =-\frac{\frac{\partial F}{\partial z}}{\frac{\partial F}{\partial y}} =\mathbf{-\frac{2x^{3} z+4xy}{3y^{2} +4xz}} \end{aligned}\]Question (d)
\[\begin{aligned} \frac{dy}{dx} & =-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} =\mathbf{-\frac{6xy^{3} +z^{2} y^{2} +4y^{3} zx^{3}}{9x^{2} y^{2} +2xz^{2} y+3y^{2} zx^{4} +2yz}}\\ \frac{dy}{dz} & =-\frac{\frac{\partial F}{\partial z}}{\frac{\partial F}{\partial y}} =\mathbf{-\frac{2xzy^{2} +y^{3} x^{4} +y^{2}}{9x^{2} y^{2} +2xz^{2} y+3y^{z} x^{4} +2yz}} \end{aligned}\]Question (e)
\[\begin{aligned} \frac{dy}{dx} & =-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} =\mathbf{-\frac{2xy\left( y^{2} -e^{x^{2}}\right)}{5y^{4} +3x^{2} y^{2} -e^{x^{2}}}} \end{aligned}\]Q8: (a) Show that the variables \(x\) and \(y\) given by \(x=\frac{s+t}{s}\), \(y=\frac{s+t}{t}\) are functionally dependent.
(b) Obtain the Jacobian \(J\) of the transformation \(s=2x+y\), \(t=x-2y\) and determine the inverse of the transformation \(J_{1}\). Confirm that \(J_{1} =J^{-1}\).
(c) Show that if \(x+y=u\) and \(y=uv\), then \(\frac{\partial ( x,y)}{\partial ( u,v)} =u\).
(d) Verify whether the functions \(u=\frac{x+y}{1-xy}\) and \(v=\tan^{-1} x+\tan^{-1} y\) are functionally dependent.
(e) If \(x=uv\), \(y=\frac{u+v}{u-v}\), find \(\frac{\partial ( u,v)}{\partial ( x,y)}\).
Solution
Question (a)
\[\begin{aligned} J & =\frac{\partial ( x,y)}{\partial ( s,t)}\\ & =\left| \begin{matrix} -\frac{t}{s^{2}} & \frac{1}{t}\\ \frac{1}{s} & -\frac{s}{t^{2}} \end{matrix}\right| \\ & =\frac{1}{st} -\frac{1}{st} =0\Longrightarrow \text{Functionally dependent} \end{aligned}\]Question (b)
\[\begin{aligned} J & =\frac{\partial ( s,t)}{\partial ( x,y)}\\ & =\left| \begin{matrix} 2 & 1\\ 1 & -2 \end{matrix}\right| \\ & =-5 \end{aligned}\] \[\begin{array}{l} x=\frac{1}{5}( 2s+t) \quad \quad y=\frac{1}{5}( s-2t)\\ \\ \begin{aligned} J_{1} & =\frac{\partial ( x,y)}{\partial ( s,t)}\\ & =\left| \begin{matrix} \frac{2}{5} & \frac{1}{5}\\ \frac{1}{5} & -\frac{2}{5} \end{matrix}\right| \\ & =-\frac{1}{5} \end{aligned} \end{array}\] \[\therefore J_{1} =J^{-1}\]Question (c)
\[\begin{array}{l} x+uv=u\Longrightarrow x=u-uv\\ \frac{dx}{du} =1-v\\ \frac{dx}{dv} =-1 \end{array}\] \[\begin{array}{l} y=uv\\ \frac{dy}{du} =v\\ \frac{dy}{dv} =u \end{array}\] \[\begin{aligned} \frac{\partial ( x,y)}{\partial ( u,v)} & =\left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{matrix}\right| \\ & =\left| \begin{matrix} 1-v & v\\ -u & u \end{matrix}\right| \\ & =u-uv+uv\\ & =u\ \end{aligned}\]Question (d)
\[\begin{aligned} \frac{\partial ( u,v)}{\partial ( x,y)} & =\left| \begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix}\right| =\left| \begin{matrix} \frac{1+y^{2}}{( 1-xy)^{2}} & \frac{1+x^{2}}{( 1-xy)^{2}}\\ \frac{1}{1+x^{2}} & \frac{1}{1+y^{2}} \end{matrix}\right| \\ & =\frac{1}{( 1-xy)^{2}} -\frac{1}{( 1-xy)^{2}}\\ & =0 \end{aligned}\]Question (e)
\[\begin{aligned} \frac{\partial ( x,y)}{\partial ( u,v)} & =\left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{matrix}\right| =\left| \begin{matrix} v & u\\ -\frac{2v}{( u-v)^{2}} & -\frac{2u}{( u-v)^{2}} \end{matrix}\right| \\ & =\frac{2uv}{( u-v)^{2}} +\frac{2uv}{( u-v)^{2}}\\ & =\frac{4uv}{( u-v)^{2}}\\ \mathbf{\frac{\partial ( u,v)}{\partial ( x,y)}} & \mathbf{=\frac{( u-v)^{2}}{4uv}} \end{aligned}\]Total Differential
Q1: Compute the total differential of \(f( x,y,z) =\ln\left(\frac{xy^{2}}{z^{3}}\right)\).
Solution
\[dF=\frac{1}{x} dx+\frac{2}{y} dy-\frac{3}{2} dz\]Q2: The period \(T\) of a simple pendulum is \(T=2\pi \sqrt{\frac{l}{g}}\), find the maximum percentage error in \(T\) due to possible errors up to 1% in \(l\) and 2% in \(g\). (Hint: \(\frac{dl}{l} =0.001\) and \(\frac{dg}{g} =0.002\))
Solution
\[\begin{aligned} \log T & =\log 2\pi +\frac{1}{2}\log l-\frac{1}{2}\log g\\ \frac{dT}{T} & =0+\frac{1}{2l} dl-\frac{1}{2g} dg\\ & =\left(\frac{1}{2}\right)\left(\frac{dl}{l}\right) -\left(\frac{1}{2}\right)\left(\frac{dg}{g}\right)\\ & =\left(\frac{1}{2}\right)( 0.001) \pm \left(\frac{1}{2}\right)( 0.002)\\ & =0.0005\pm 0.001 \end{aligned}\] \[\therefore Max\ error\ =\ 1.5\%\]Q3: Compute an approximate value of \(( 1.04)^{3.01}\).
Solution
Let \(f( x,y) =x^{y}\).
\[\frac{\partial f}{\partial x} =yx^{y-1} \quad \quad \frac{\partial f}{\partial y} =x^{y}\log x\] \[x=1,\ dx=0.04\quad \quad y=3,\ dy=0.01\] \[\begin{aligned} df & =\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy\\ & =yx^{y-1} \ dx+x^{y}\log x\ dy\ \\ & =3( 1)^{3-1}( 0.004) +1^{3}\log( 1) \ ( 0.01)\\ & =0.12 \end{aligned}\] \[\begin{aligned} ( 1.04)^{3.01} & =f( 1,3) +df\\ & =1+0.12\\ & =\mathbf{1.12} \end{aligned}\]Q4: Suppose one is given a triangle where the angle at one vertex is \(\theta\) and the lengths of the two sides adjacent to that vertex are \(b\) and \(c\). Then the well-known formula for the area of the triangle as
\[\text{Area} =S=\frac{1}{2} bc\sin \theta\]Suppose we measure:
\[\begin{aligned} b & =4.00\ m\ \pm \ 0.005\ m\\ c & =3.00\ m\ \pm \ 0.005\ m\\ \theta & =\frac{\pi }{6} \ \pm \ 0.01\ radians \end{aligned}\]Find \(S\) and estimate the percentage error.
Solution
Area, \(S=\frac{1}{2}( 4)( 3)\sin\left(\frac{\pi }{6}\right) =3\)
\[\begin{aligned} \frac{\partial S}{\partial b} & =\frac{1}{2} c\sin \theta \\ \frac{\partial S}{\partial c} & =\frac{1}{2} b\sin \theta \\ \frac{\partial S}{\partial \theta } & =\frac{1}{2} bc\cos \theta \end{aligned}\]The error estimate,
\[\begin{aligned} \text{Max Error} & \leqslant \left| \frac{\partial S}{\partial b} db\right| +\left| \frac{\partial S}{\partial c} dc\right| +\left| \frac{\partial S}{\partial \theta } d\theta \right| \\ & =\left(\frac{1}{2} \times 3\times \frac{1}{2}\right) \times 0.005+\left(\frac{1}{2} \times 4\times \frac{1}{2}\right) \times 0.005+\left(\frac{1}{2} \times 4\times 3\times \frac{\sqrt{3}}{2}\right) \times 0.01\\ & =0.06 \end{aligned}\]The relative error is at most \(\frac{0.06}{3} =0.02\).
\(\Longrightarrow \text{Percentage Error} =2\%\).
Tangent planes and normal to surfaces in three dimensions
Find the equations of the tangent plane and normal line to the following surfaces at the points indicated:
Q1: \(x^{2} +2y^{2} +3z^{2} =6\) at \((1,1,1)\).
Solution
Partial derivation yields:
\[\frac{\partial f}{\partial x} =2x\quad \quad \frac{\partial f}{\partial y} =4y\quad \quad \frac{\partial f}{\partial z} =6z\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 1,1,1)} =2\quad \quad \frac{\partial f}{\partial y}_{( 1,1,1)} =4\quad \quad \frac{\partial f}{\partial z}_{( 1,1,1)} =6\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 2( x-1) +4( y-1) +6( z-1) & =0\\ 2x-2+4y-4+6z-6 & =0\\ \mathbf{x+2y+3z} & \mathbf{=6} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \frac{x-1}{2} & =\frac{y-1}{4} =\frac{z-1}{6}\\ \mathbf{\frac{x-1}{1}} & \mathbf{=\frac{y-1}{2} =\frac{z-1}{3}} \end{aligned}\]Q2: \(2x^{2} +y^{2} -z^{2} =-3\) at \((1,2,3)\).
Solution
Partial derivation yields:
\[\frac{\partial f}{\partial x} =4x\quad \quad \frac{\partial f}{\partial y} =2y\quad \quad \frac{\partial f}{\partial z} =-2z\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 1,2,3)} =4\quad \quad \frac{\partial f}{\partial y}_{( 1,2,3)} =4\quad \quad \frac{\partial f}{\partial z}_{( 1,2,3)} =-6\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 4( x-1) +4( y-2) -6( z-3) & =0\\ 4x-4+4y-8-6z+18 & =0\\ 4x+4y-6z & =-6\\ 2\mathbf{x+2y-3z} & \mathbf{=-3} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \frac{x-1}{4} & =\frac{y-2}{4} =\frac{z-3}{-6}\\ \mathbf{\frac{x-1}{2}} & \mathbf{=\frac{y-2}{2} =\frac{z-3}{-3}} \end{aligned}\]Q3: \(x^{2} +y^{2} -z=1\) at \((1,2,4)\).
Solution
Partial derivation yields:
\[\frac{\partial f}{\partial x} =2x\quad \quad \frac{\partial f}{\partial y} =2y\quad \quad \frac{\partial f}{\partial z} =-1\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 1,2,4)} =2\quad \quad \frac{\partial f}{\partial y}_{( 1,2,4)} =4\quad \quad \frac{\partial f}{\partial z}_{( 1,2,4)} =-1\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 2( x-1) +4( y-2) -1( z-4) & =0\\ 2x-2+4y-8-z+4 & =0\\ \mathbf{2x+4y-z} & \mathbf{=6} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \mathbf{\frac{x-1}{2}} & \mathbf{=\frac{y-2}{4} =\frac{z-4}{-1}} \end{aligned}\]Q4: \(\ln\left(\frac{x}{y}\right) -z^{2}( x-2y) -3z=3\) at \((4,2,-1)\).
Solution
Partial derivation yields:
\[\frac{\partial f}{\partial x} =\frac{1}{x} -z^{2} \quad \quad \frac{\partial f}{\partial y} =2z^{2} -\frac{1}{y} \quad \quad \frac{\partial f}{\partial z} =-2z( x-2y) -3\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 4,2,-1)} =-\frac{3}{4} \quad \quad \frac{\partial f}{\partial y}_{( 4,2,-1)} =\frac{3}{2} \quad \quad \frac{\partial f}{\partial z}_{( 4,2,-1)} =-3\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ -\frac{3}{4}( x-4) +\frac{3}{2}( y-2) -3( z+1) & =0\\ -\frac{3}{4} x+3+\frac{3}{2} y-3-3z-3 & =0\\ \mathbf{-\frac{3}{4} x+\frac{3}{2} y-3z} & \mathbf{=3} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \mathbf{\frac{x-4}{-\frac{3}{4}}} & \mathbf{=\frac{y-2}{\frac{3}{2}} =\frac{z+1}{-3}} \end{aligned}\]Q5: \(x^{3} z+z^{3} x-2yz=0\) at \((1,1,1)\).
Solution
Partial derivation yields:
\[\frac{\partial f}{\partial x} =3x^{2} z+z^{3} \quad \quad \frac{\partial f}{\partial y} =-2z\quad \quad \frac{\partial f}{\partial z} =x^{3} +3z^{2} x-2y\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 1,1,1)} =4\quad \quad \frac{\partial f}{\partial y}_{( 1,1,1)} =-2\quad \quad \frac{\partial f}{\partial z}_{( 1,1,1)} =2\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 4( x-1) -2( y-1) +2( z-1) & =0\\ 4x-4-2y+2+2z-2 & =0\\ 4x-2y+2z & =4\\ \mathbf{2x-y+z} & \mathbf{=2} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \frac{x-1}{4} & =\frac{y-1}{-2} =\frac{z-1}{2}\\ \mathbf{\frac{x-1}{2}} & \mathbf{=\frac{y-1}{-1} =\frac{z-1}{1}} \end{aligned}\]Q6: \(z=5+( x-1)^{2} +( y+2)^{2}\) at \((2,0,10)\).
Solution
\[z=5+( x-1)^{2} +( y+2)^{2} \Longrightarrow ( x-1)^{2} +( y+2)^{2} -z=5\]Partial derivation yields:
\[\frac{\partial f}{\partial x} =2( x-1) \quad \quad \frac{\partial f}{\partial y} =2( y+2) \quad \quad \frac{\partial f}{\partial z} =-1\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 2,0,10)} =2\quad \quad \frac{\partial f}{\partial y}_{( 2,0,10)} =4\quad \quad \frac{\partial f}{\partial z}_{( 2,0,10)} =-1\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 2( x-2) +4( y) -( z-10) & =0\\ 2x-4+4y-z+10 & =0\\ \mathbf{2x+4y-z} & \mathbf{=-6} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \mathbf{\frac{x-2}{2}} & \mathbf{=\frac{y}{4} =\frac{z-10}{-1}} \end{aligned}\]Q7: \(\frac{x^{2}}{12} +\frac{y^{2}}{6} +\frac{z^{2}}{4} =1\) at \((1,2,1)\).
Solution
Partial derivation yields:
\[\frac{\partial f}{\partial x} =\frac{x}{6} \quad \quad \frac{\partial f}{\partial y} =\frac{y}{3} \quad \quad \frac{\partial f}{\partial z} =\frac{z}{2}\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 1,2,1)} =\frac{1}{6} \quad \quad \frac{\partial f}{\partial y}_{( 1,2,1)} =\frac{2}{3} \quad \quad \frac{\partial f}{\partial z}_{( 1,2,1)} =\frac{1}{2}\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ \frac{1}{6}( x-1) +\frac{2}{3}( y-2) +\frac{1}{2}( z-1) & =0\\ \frac{1}{6} x-\frac{1}{6} +\frac{2}{3} y-\frac{4}{3} +\frac{1}{2} z-\frac{1}{2} & =0\\ \frac{1}{6} x+\frac{4}{6} y+\frac{3}{6} z-\frac{1}{6} -\frac{8}{6} -\frac{3}{6} & =0\\ \frac{1}{6} x+\frac{4}{6} y+\frac{3}{6} z-2 & =0\\ \mathbf{x+4y+3z} & \mathbf{=12} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \frac{x-1}{\frac{1}{6}} & =\frac{y-2}{\frac{2}{3}} =\frac{z-1}{\frac{1}{2}}\\ \frac{x-1}{\frac{1}{6}} & =\frac{y-2}{\frac{4}{6}} =\frac{z-1}{\frac{3}{6}}\\ \mathbf{\frac{x-1}{1}} & \mathbf{=\frac{y-2}{4} =\frac{z-1}{3}} \end{aligned}\]Q8: \(ze^{x} +e^{z+1} +xy+y=3\) at \((0,3,-1)\).
Solution
Partial derivation yields:
\[\frac{\partial f}{\partial x} =ze^{x} +y\quad \quad \frac{\partial f}{\partial y} =x+1\quad \quad \frac{\partial f}{\partial z} =e^{x} +e^{z+1}\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 0,3,-1)} =2\quad \quad \frac{\partial f}{\partial y}_{( 0,3,-1)} =1\quad \quad \frac{\partial f}{\partial z}_{( 0,3,-1)} =2\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 2( x) +( y-3) +2( z+1) & =0\\ 2x+y-3+2z+2 & =0\\ \mathbf{2x+y+2z} & \mathbf{=1} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \mathbf{\frac{x}{2}} & \mathbf{=\frac{y-3}{1} =\frac{z+1}{2}} \end{aligned}\]Q9: \(z^{2} =7-x^{2} -2y^{2}\) at \((1,1,2)\).
Solution
\[z^{2} =7-x^{2} -2y^{2} \Longrightarrow x^{2} +2y^{2} +z^{2} =7\]Partial derivation yields:
\[\frac{\partial f}{\partial x} =2x\quad \quad \frac{\partial f}{\partial y} =4y\quad \quad \frac{\partial f}{\partial z} =2z\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{( 1,1,2)} =2\quad \quad \frac{\partial f}{\partial y}_{( 1,1,2)} =4\quad \quad \frac{\partial f}{\partial z}_{( 1,1,2)} =4\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 2( x-1) +4( y-1) +4( z-2) & =0\\ 2x-2+4y-4+4z-8 & =0\\ 2x+4y+4z-14 & =0\\ \mathbf{x+2y+2z} & \mathbf{=7} \end{aligned}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \frac{x-1}{2} & =\frac{y-1}{4} =\frac{z-2}{4}\\ \mathbf{\frac{x-1}{1}} & \mathbf{=\frac{y-1}{2} =\frac{z-2}{2}} \end{aligned}\]Q10: \(z=\frac{x^{2}}{2p} +\frac{y^{2}}{2q}\) at the point \(\left( 2p^{2} ,2q^{2} ,2p^{3} +2q^{3}\right)\).
Solution
\[z=\frac{x^{2}}{2p} +\frac{y^{2}}{2q} \Longrightarrow \frac{x^{2}}{2p} +\frac{y^{2}}{2q} -z=0\]Partial derivation yields:
\[\frac{\partial f}{\partial x} =\frac{x}{p} \quad \quad \frac{\partial f}{\partial y} =\frac{y}{q} \quad \quad \frac{\partial f}{\partial z} =-1\] \[\Longrightarrow \ \frac{\partial f}{\partial x}_{\left( 2p^{2} ,2q^{2} ,2p^{3} +2q^{3}\right)} =2p\quad \quad \frac{\partial f}{\partial y}_{\left( 2p^{2} ,2q^{2} ,2p^{3} +2q^{3}\right)} =2q\quad \quad \frac{\partial f}{\partial z}_{\left( 2p^{2} ,2q^{2} ,2p^{3} +2q^{3}\right)} =-1\]Tangent plane:
\[\begin{aligned} ( x-x_{o})\left(\frac{\partial f}{\partial x}\right)_{0} +( y-y_{0})\left(\frac{\partial f}{\partial y}\right)_{0} +( z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{0} & =0\\ 2p\left( x-2p^{2}\right) +2q\left( y-2q^{2}\right) -1\left( z-\left( 2p^{3} +2q^{3}\right)\right) & =0\\ 2px-4p^{3} +2qy-4q^{3} -z+2p^{3} +2q^{3} & =0\\ 2px+2qy-z-2p^{3} -2q^{3} & =0 \end{aligned}\\ \\ \mathbf{\begin{aligned} z & =2px+2qy-2p^{3} -2q^{3} \end{aligned}}\]Normal line:
\[\begin{aligned} \frac{x-x_{0}}{\left(\frac{\partial f}{\partial x}\right)_{0}} & =\frac{y-y_{0}}{\left(\frac{\partial f}{\partial y}\right)_{0}} =\frac{z-z_{0}}{\left(\frac{\partial f}{\partial z}\right)_{0}}\\ \mathbf{\frac{x-2p^{2}}{2p}} & \mathbf{=\frac{y-2q^{2}}{2q} =\frac{z-2p^{3} -2q^{3}}{-1}} \end{aligned}\]