Calculus Volume 3

# Key Equations

Calculus Volume 3Key Equations

### Key Equations

 Vertical trace $f(a,y)=zf(a,y)=z$ for $x=ax=a$ or $f(x,b)=zf(x,b)=z$ for $y=by=b$ Level surface of a function of three variables $f(x,y,z)=cf(x,y,z)=c$
 Partial derivative of $ff$ with respect to $xx$ $∂f∂x=limh→0f(x+h,y)−f(x,y)h∂f∂x=limh→0f(x+h,y)−f(x,y)h$ Partial derivative of $ff$ with respect to $yy$ $∂f∂y=limk→0f(x,y+k)−f(x,y)k∂f∂y=limk→0f(x,y+k)−f(x,y)k$
 Tangent plane $z=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)z=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)$ Linear approximation $L(x,y)=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)L(x,y)=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)$ Total differential $dz=fx(x0,y0)dx+fy(x0,y0)dy.dz=fx(x0,y0)dx+fy(x0,y0)dy.$ Differentiability (two variables) $f(x,y)=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)+E(x,y),f(x,y)=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)+E(x,y),$ where the error term $EE$ satisfies $lim(x,y)→(x0,y0)E(x,y)(x−x0)2+(y−y0)2=0.lim(x,y)→(x0,y0)E(x,y)(x−x0)2+(y−y0)2=0.$ Differentiability (three variables) $f(x,y)=f(x0,y0,z0)+fx(x0,y0,z0)(x−x0)+fy(x0,y0,z0)(y−y0)+fz(x0,y0,z0)(z−z0)+E(x,y,z),f(x,y)=f(x0,y0,z0)+fx(x0,y0,z0)(x−x0)+fy(x0,y0,z0)(y−y0)+fz(x0,y0,z0)(z−z0)+E(x,y,z),$ where the error term $EE$ satisfies $lim(x,y,z)→(x0,y0,z0)E(x,y,z)(x−x0)2+(y−y0)2+(z−z0)2=0.lim(x,y,z)→(x0,y0,z0)E(x,y,z)(x−x0)2+(y−y0)2+(z−z0)2=0.$
 Chain rule, one independent variable $dzdt=∂z∂x·dxdt+∂z∂y·dydtdzdt=∂z∂x·dxdt+∂z∂y·dydt$ Chain rule, two independent variables $dzdu=∂z∂x∂x∂u+∂z∂y∂x∂udzdu=∂z∂x∂x∂u+∂z∂y∂x∂u$ $dzdv=∂z∂x∂x∂v+∂z∂y∂y∂vdzdv=∂z∂x∂x∂v+∂z∂y∂y∂v$ Generalized chain rule $∂w∂tj=∂w∂x1∂x1∂tj+∂w∂x2∂x1∂tj+⋯+∂w∂xm∂xm∂tj∂w∂tj=∂w∂x1∂x1∂tj+∂w∂x2∂x1∂tj+⋯+∂w∂xm∂xm∂tj$
 directional derivative (two dimensions) $Duf(a,b)=limh→0f(a+hcosθ,b+hsinθ)−f(a,b)hDuf(a,b)=limh→0f(a+hcosθ,b+hsinθ)−f(a,b)h$ or $Duf(x,y)=fx(x,y)cosθ+fy(x,y)sinθDuf(x,y)=fx(x,y)cosθ+fy(x,y)sinθ$ gradient (two dimensions) $∇f(x,y)=fx(x,y)i+fy(x,y)j∇f(x,y)=fx(x,y)i+fy(x,y)j$ gradient (three dimensions) $∇f(x,y,z)=fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k∇f(x,y,z)=fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k$ directional derivative (three dimensions) $Duf(x,y,z)=∇f(x,y,z)·u=fx(x,y,z)cosα+fy(x,y,z)cosβ+fx(x,y,z)cosγDuf(x,y,z)=∇f(x,y,z)·u=fx(x,y,z)cosα+fy(x,y,z)cosβ+fx(x,y,z)cosγ$
 Discriminant $D=fxx(x0,y0)fyy(x0,y0)−(fxy(x0,y0))2D=fxx(x0,y0)fyy(x0,y0)−(fxy(x0,y0))2$
 Method of Lagrange multipliers, one constraint $∇f(x0,y0)=λ∇g(x0,y0)g(x0,y0)=0∇f(x0,y0)=λ∇g(x0,y0)g(x0,y0)=0$ Method of Lagrange multipliers, two constraints $∇f(x0,y0,z0)=λ1∇g(x0,y0,z0)+λ2∇h(x0,y0,z0)g(x0,y0,z0)=0h(x0,y0,z0)=0∇f(x0,y0,z0)=λ1∇g(x0,y0,z0)+λ2∇h(x0,y0,z0)g(x0,y0,z0)=0h(x0,y0,z0)=0$
Order a print copy

As an Amazon Associate we earn from qualifying purchases.