Wize University Calculus 2 Textbook > Multivariable Functions

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Double Integration (rectangular)Next Section

Multivariable Functions Quiz

Calculate ∂f∂x\frac{\partial f}{\partial x}∂x∂f​, ∂f∂y\frac{\partial f}{\partial y}∂y∂f​, ∂2f∂x2\frac{\partial^2f}{\partial x^2}∂x2∂2f​, ∂2f∂y2\frac{\partial^2f}{\partial y^2}∂y2∂2f​, ∂f∂x∂y\frac{\partial f}{\partial x\partial y}∂x∂y∂f​ for f(x,y)=cos⁡(xy)f(x,y)=\cos(xy)f(x,y)=cos(xy).

fx=sin⁡(xy)yfxx=cos⁡(xy)y2fy=−sin⁡(xy)xfyy=−cos⁡(xy)x2fxy=−cos⁡(xy)xy+(−sin⁡(xy))\begin{array}{l}
\begin{array}{ll}
f_x=\sin(xy)y & f_{xx}=\cos(xy)y^2\\
f_y=-\sin(xy)x & f_{yy}=-\cos(xy)x^2
\end{array}\\
f_{xy}=-\cos(xy)xy+(-\sin(xy))
\end{array}
fx​=sin(xy)yfy​=−sin(xy)x​fxx​=cos(xy)y2fyy​=−cos(xy)x2​fxy​=−cos(xy)xy+(−sin(xy))​

fx=−sin⁡(xy)xfxx=−cos⁡(xy)x2fy=−sin⁡(xy)yfyy=−cos⁡(xy)y2fxy=−cos⁡(xy)xy+(−sin⁡(xy))\begin{array}{l}
\begin{array}{ll}
f_x=-\sin(xy)x & f_{xx}=-\cos(xy)x^2\\
f_y=-\sin(xy)y & f_{yy}=-\cos(xy)y^2
\end{array}\\
f_{xy}=-\cos(xy)xy+(-\sin(xy))
\end{array}fx​=−sin(xy)xfy​=−sin(xy)y​fxx​=−cos(xy)x2fyy​=−cos(xy)y2​fxy​=−cos(xy)xy+(−sin(xy))​

fx=−sin⁡(xy)yfxx=−cos⁡(xy)y2fy=−sin⁡(xy)xfyy=−cos⁡(xy)x2fxy=−cos⁡(xy)xy+(−sin⁡(xy))\begin{array}{l}
\begin{array}{ll}
f_x=-\sin(xy)y & f_{xx}=-\cos(xy)y^2\\
f_y=-\sin(xy)x & f_{yy}=-\cos(xy)x^2
\end{array}\\
f_{xy}=-\cos(xy)xy+(-\sin(xy))
\end{array}
fx​=−sin(xy)yfy​=−sin(xy)x​fxx​=−cos(xy)y2fyy​=−cos(xy)x2​fxy​=−cos(xy)xy+(−sin(xy))​

fx=−sin⁡(xy)yfxx=−cos⁡(xy)y2fy=−sin⁡(xy)xfyy=−cos⁡(xy)x2fxy=cos⁡(xy)xy+sin⁡(xy)\begin{array}{l}
\begin{array}{ll}
f_x=-\sin(xy)y & f_{xx}=-\cos(xy)y^2\\
f_y=-\sin(xy)x & f_{yy}=-\cos(xy)x^2
\end{array}\\
f_{xy}=\cos(xy)xy+\sin(xy)
\end{array}
fx​=−sin(xy)yfy​=−sin(xy)x​fxx​=−cos(xy)y2fyy​=−cos(xy)x2​fxy​=cos(xy)xy+sin(xy)​

None of the above

I don't know

Given the implicit relation 4x3+5y2−2z2+3xy−4xz+yz=3,4x^3+5y^2-2z^2+3xy-4xz+yz=3,4x3+5y2−2z2+3xy−4xz+yz=3, calculate ∂z∂x,∂z∂y, and ∂2z∂x2\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \text{ and }\frac{\partial^2 z}{\partial x^2}∂x∂z​,∂y∂z​, and ∂x2∂2z​

∂z∂x=12x2+3y−4z4z+4x−y\displaystyle \frac{\partial z}{\partial x}=\frac{12x^2+3y-4z}{4z+4x-y}∂x∂z​=4z+4x−y12x2+3y−4z​
∂z∂y=10y+3x+z4z+4x−y\displaystyle \frac{\partial z}{\partial y}=\frac{10y+3x+z}{4z+4x-y}∂y∂z​=4z+4x−y10y+3x+z​
∂z∂x=6x\displaystyle \frac{\partial z}{\partial x}=6x∂x∂z​=6x

∂z∂x=12x2+3y−4z4z+4x−y\displaystyle \frac{\partial z}{\partial x}=\frac{12x^2+3y-4z}{4z+4x-y}∂x∂z​=4z+4x−y12x2+3y−4z​
∂z∂y=10y+3x+z4z+4x−y\displaystyle \frac{\partial z}{\partial y}=\frac{10y+3x+z}{4z+4x-y}∂y∂z​=4z+4x−y10y+3x+z​
∂2z∂x2=(24x)(4z+4x−y)−(12x2+3y−4z)(4)(4z+4x−y)2\displaystyle \frac{\partial^2z}{\partial x^2}=\frac{\left(24x\right)\left(4z+4x-y\right)-\left(12x^2+3y-4z\right)\left(4\right)}{\left(4z+4x-y\right)^2}∂x2∂2z​=(4z+4x−y)2(24x)(4z+4x−y)−(12x2+3y−4z)(4)​

∂z∂x=12x2+3y−4z4z+4x−y\displaystyle \frac{\partial z}{\partial x}=\frac{12x^2+3y-4z}{4z+4x-y}∂x∂z​=4z+4x−y12x2+3y−4z​
∂z∂y=10y+3x+z4z+4x−y\displaystyle \frac{\partial z}{\partial y}=\frac{10y+3x+z}{4z+4x-y}∂y∂z​=4z+4x−y10y+3x+z​
∂2z∂x2=(24x−4 ∂z∂x)(4z+4x−y)−(12x2+3y−4z)(4 ∂z∂x+4)(4z+4x−y)2\displaystyle \frac{\partial^2z}{\partial x^2}=\frac{\left(24x-4\ \frac{\partial z}{\partial x}\right)\left(4z+4x-y\right)-\left(12x^2+3y-4z\right)\left(4\ \frac{\partial z}{\partial x}+4\right)}{\left(4z+4x-y\right)^2}∂x2∂2z​=(4z+4x−y)2(24x−4 ∂x∂z​)(4z+4x−y)−(12x2+3y−4z)(4 ∂x∂z​+4)​

None of the above

I don't know

Let f(x,y)=xx2+y2.f(x,y)=\frac{x}{\sqrt{x^2+y^2}}.f(x,y)=x2+y2​x​. Find ∂f∂x(x,y).\frac{\partial f}{\partial x}\left(x,y\right).∂x∂f​(x,y).

Answer

I don't know

Let f(x,y,z)=exy2z3.f(x,y,z)=e^{xy^2z^3}.f(x,y,z)=exy2z3. Find fzyx(x,y,z).f_{zyx}\left(x,y,z\right).fzyx​(x,y,z).

Answer

I don't know

Suppose that f(x,y,z)=x3+y2+3z2f\left(x,y,z\right)=x^3+y^2+3z^2f(x,y,z)=x3+y2+3z2 and x=arcsin⁡(t)x=\arcsin\left(t\right)x=arcsin(t), y=2ety=2e^ty=2et, z=ln⁡(1+t2)z=\ln\left(1+t^2\right)z=ln(1+t2). Find ∂f∂t\displaystyle \frac{\partial f}{\partial t}∂t∂f​ at t=0t=0t=0.

Answer

I don't know

If w=x2y2z2w = x^2 y^2 z^2w=x2y2z2 where x=1/(t2+s2), y=t−7, z=1/t2+1/s2,x = 1/(t^2  + s^2),\ y = t^{-7},\ z = 1/t^2  + 1/s^2,x=1/(t2+s2), y=t−7, z=1/t2+1/s2, find ∂w∂t\frac{\partial w}{\partial t}∂t∂w​ at (s, t) = (2, 1). 

Answer

I don't know

If f(x,y)=x+arccos⁡(xy)2+y,f\left(x,y\right)=\frac{x+\arccos\left(xy\right)}{2+y},f(x,y)=2+yx+arccos(xy)​, then what is the value of f(3, 12)?f\left(\sqrt{3,}\ \frac{1}{2}\right)?f(3,​ 21​)? 

(a) 3+π\sqrt{3}+\pi3​+π

(b)63+π15\frac{6\sqrt{3}+\pi}{15}1563​+π​

(c)235+2π15\frac{2\sqrt{3}}{5}+\frac{2\pi}{15}523​​+152π​

(d)3+π6(2+3)\frac{3+\pi}{6\left(2+\sqrt{3}\right)}6(2+3​)3+π​

(e)14+23+π6+33\frac{1}{4+2\sqrt{3}}+\frac{\pi}{6+3\sqrt{3}}4+23​1​+6+33​π​

I don't know

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State and sketch the the domain of f(x,y)=x−3y+ln⁡(x2+y2−3)f(x,y)=\sqrt{x-3y}+\ln\left(x^2+y^2-3\right)f(x,y)=x−3y​+ln(x2+y2−3).

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For the surface described by y2=x2+z2y^2=x^2+z^2y2=x2+z2, find traces in the planes x=Kx=Kx=K, y=Ky=Ky=K, z=Kz=Kz=K. Then identify the surface and sketch it in 3D.

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Which of the following is the tangent plane to 0=2xy2+yz2−x2yz0=2xy^2+yz^2-x^2yz0=2xy2+yz2−x2yz at the point (1,1,−1)(1,1,-1)(1,1,−1). 

2(x−1)+6(y−1)+0(z+1)=02(x-1)+6(y-1)+0(z+1)=02(x−1)+6(y−1)+0(z+1)=0

4(x−1)+6(y−1)−3(z+1)=04(x-1)+6(y-1)-3(z+1)=04(x−1)+6(y−1)−3(z+1)=0

z=−1+4(x−1)+6(y−1)z=-1+4(x-1)+6(y-1)z=−1+4(x−1)+6(y−1)

z=−1+43(x−1)+6(y−1)z=-1+\frac{4}{3}(x-1)+6(y-1)z=−1+34​(x−1)+6(y−1)

z=−3+4(x−1)+6(y−1)z=-3+4(x-1)+6(y-1)z=−3+4(x−1)+6(y−1)

I don't know

Find the maximum value of f(x,y)=x+2yf\left(x,y\right)=x+2yf(x,y)=x+2y subject to the constraint x 2+2y=10x\ ^2+2y=10x 2+2y=10

Suppose we want to minimize the cost of producing a cylindrical can with radius rrr and height hhh. The volume of the can must be 350cm3 and the material used to produce the can costs $0.5/cm2

Which of the following models this problem as an optimization problem?

Minimize S=0.5(πr2+2πrh)S=0.5\left(\pi r^2+2\pi rh\right)S=0.5(πr2+2πrh) where πr2h=350\pi r^2h=350πr2h=350

Minimize S=0.5(2πr2+2πrh)S=0.5\left(2\pi r^2+2\pi rh\right)S=0.5(2πr2+2πrh) where πr2h=350\pi r^2h=350πr2h=350

Minimize S=0.5(πr2+πrh)S=0.5\left(\pi r^2+\pi rh\right)S=0.5(πr2+πrh) where πrh=350\pi rh=350πrh=350

Minimize S=0.5(2πr2 h)S=0.5\left(2\pi r^{2\ }h\right)S=0.5(2πr2 h) where πr2h=350\pi r^2h=350πr2h=350

Minimize S=0.5(πr2+πh+r2h)S=0.5\left(\pi r^2+\pi h+r^2h\right)S=0.5(πr2+πh+r2h) where πrh=350\pi rh=350πrh=350

I don't know

Wize University Calculus 2 Textbook > Multivariable Functions

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Multivariable Functions Quiz