Calculus 3 Worksheet 1

Questions

1. Find partial derivative $\partial f/\partial x$ and $\partial f/\partial y$ of the following function

   • $f(x,y)=\ln (xy)\tan (xy)$.
   • $f(x,y)=u^2-v^2$, $u=\sec (x)$, and $v=\tan (y)$.

2. Let $F(x,y)$ be a differtiable function. Let $F_x(x,y)=G(x,y)$ and $F_y(x,y)=H(x,y)$.

   • Find $u$ and $v$ such that $F(u,v)=F(x^3+1,2x)=f(x)$.
   • Use chain rule to find $f^{\prime }(x)$, and express the answer in terms of $G$ and $H$.
   • Use the same reasoning to find the derivative of $f(x)=F(x,x)$, and express the derivative in terms of $G$ and $H$.

3. The gradient $\mathrm{\nabla }F(a,b)$ of a function $F(x,y)$ at $(a,b)$ is a vector defined by $(F_x(a,b),F_y(a,b))$. The gradient of a function can utilize to find the directional derivative $F_{(u,v)}$ of $F$ along a vector $(u,v)$. It is [F_{(u,v)}(a,b)=\nabla F(a,b)\cdot \frac{(u,v)}{|(u,v)|}] where $|(u,v)|=\sqrt{u^2+v^2}$ is the length of the vector $(u,v)$. Use the above introduction to calculate

   • The gradient of $F(x,y)=x^2+y^2$ at $(1,1)$ and $(0,0)$.
   • Observe that how the length of the gradient changes with respect to the point.
   • Find the directional derivative of $F$ at $(1,1)$ along the vector (3,4).
   • Confirm that the largest directional derivative of $F$ at $(1,1)$ is along the vector $(1,1)$.

Answers

1. • $f_x=\frac{1}{x}\tan (xy)+\ln (xy){\sec }^2(xy)y$, $f_y=\frac{1}{y}\tan (xy)+\ln (xy){\sec }^2(xy)x$.
   • $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=2(\sec (x))\sec (x)\tan (x)-2(\tan (x))\cdot 0=2{\sec }^2(x)\tan (x).$ Similarly, $\frac{\partial f}{\partial y}=-2\tan (y){\sec }^2(y)$.
2. • $u=x^3+1$ and $v=2x$.
   • $f^{\prime }(x)=\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x}=G(x^3+1,2x)\cdot 3x^2+H(x^3+1,2x)\cdot 2$.
   • We set $u=x$ and $v=x$, so $f^{\prime }(x)=G(x,x)+H(x,x)$.
3. The partial derivatives of $F$ are $\partial F/\partial x=2x$ and $\partial F/\partial y=2y$.
   • $\mathrm{\nabla }F(1,1)=(2,2)$.
   • $\mathrm{\nabla }F(1,1)\frac{\overrightarrow{(3,4)}}{|(3,4)|}=(\frac{6}{5},\frac{8}{5})$.
   • The largest directional derivative of $F$ at $(1,1)$ is attained along any vector that points to the same direction as the gradient $(2,2)$, including $(1,1)$.
   • The length of the gradient at any point $(a,b)$ is $|\mathrm{\nabla }F(a,b)|=2\sqrt{a^2+b^2}=2|(a,b)-(0,0)|$. We observe that the length of the gradient is twice the distance between the point and the origin.