1. Problem 17: Find and classify critical points of $f(x,y)=x^3+3xy^2-15x^2-15y^2+72x$. 2. Compute partial derivatives and set gradient to zero. 3. $f_x=3x^2+3y^2-30x+72$. 4. $f_y=6xy-30y=6y(x-5)$. 5. From $f_y=0$ we get $y=0$ or $x=5$. 6. If $y=0$ then $f_x=3x^2-30x+72=3(x-4)(x-6)$ giving $x=4,6$. 7. If $x=5$ then $f_x=3y^2-3=3(y^2-1)$ giving $y=\pm 1$. 8. Hessian entries are $f_{xx}=6x-30$, $f_{yy}=6x-30$, $f_{xy}=6y$. 9. At $(4,0)$ we have $f_{xx}=-6$, $D=36>0$, so a local maximum with $f(4,0)=112$. 10. At $(6,0)$ we have $f_{xx}=6$, $D=36>0$, so a local minimum with $f(6,0)=108$. 11. At $(5,\pm 1)$ we have $D=-36<0$, so these are saddle points. 1. Problem 18: Examine extrema of $f(x,y)=x^4+y^4-2(x-y)^2$. 2. Compute partials $f_x=4x^3-4(x-y)$ and $f_y=4y^3+4(x-y)$. 3. Setting them to zero and simplifying gives system $x^3-x+y=0$ and $y^3-y+x=0$. 4. Subtracting gives $(x-y)(x^2+xy+y^2-2)=0$, so either $x=y$ or $x^2+xy+y^2=2$. 5. If $x=y$ then $x^3=0$ so $(0,0)$ is a critical point. 6. Adding the original equations gives $x^3+y^3=0$, so either $x=-y$ leading to $x^2=2$ and points $(\sqrt{2},-\sqrt{2})$ and $(-\sqrt{2},\sqrt{2})$. 7. Hessian: $f_{xx}=12x^2-4$, $f_{yy}=12y^2-4$, $f_{xy}=4$, so $D=(12x^2-4)(12y^2-4)-16$. 8. At $(0,0)$ we get $D=0$ and quadratic term is $-2(x-y)^2$, so $(0,0)$ is a local maximum with $f(0,0)=0$. 9. At $(\pm\sqrt{2},\mp\sqrt{2})$ we have $f_{xx}=20$, $D=384>0$ and $f_{xx}>0$, so these are local minima with $f= -8$. 1. Problem 19: Expand $e^x\cos y$ about $(1,\pi/4)$ up to second degree terms. 2. Let $h=x-1$ and $k=y-\pi/4$ so $e^x\cos y=e^{1+h}\cos(\pi/4+k)=\dfrac{e}{\sqrt{2}}e^{h}(\cos k-\sin k)$. 3. Use $e^{h}=1+h+\tfrac{h^2}{2}+\cdots$, $\cos k=1-\tfrac{k^2}{2}+\cdots$, $\sin k=k+\cdots$ and keep terms up to total degree $2$. 4. Compute $(\cos k-\sin k)\approx 1-k-\tfrac{k^2}{2}$ and multiply by $e^{h}\approx 1+h+\tfrac{h^2}{2}$ keeping degrees $\le 2$. 5. The result up to second degree is $e^x\cos y\approx \dfrac{e}{\sqrt{2}}\bigl(1+h+\tfrac{h^2}{2}-k-hk-\tfrac{k^2}{2}\bigr)$. 6. Replacing $h=x-1$ and $k=y-\pi/4$ gives the expansion. 1. Problem 20: Expand $xy^2+\cos(xy)$ about $(1,\pi/2)$ up to second order. 2. Let $h=x-1$ and $k=y-\pi/2$ and compute function and derivatives at the point where $xy=\pi/2$ and $\sin(xy)=1$, $\cos(xy)=0$. 3. $f(1,\pi/2)=\pi^2/4$, $f_x(1,\pi/2)=\pi^2/4-\pi/2$, $f_y(1,\pi/2)=\pi-1$. 4. Second derivatives at the point are $f_{xx}=0$, $f_{xy}=\pi-1$, $f_{yy}=2$. 5. The Taylor polynomial up to second order is 6. $f\approx \dfrac{\pi^2}{4}+(\dfrac{\pi^2}{4}-\dfrac{\pi}{2})(x-1)+(\pi-1)(y-\tfrac{\pi}{2})+(\pi-1)(x-1)(y-\tfrac{\pi}{2})+(y-\tfrac{\pi}{2})^2$. 1. Problem 21: Expand $y^x$ about $(1,1)$ up to second degree terms and find its reciprocal at $(1,1)$. 2. Put $x=1+h$, $y=1+k$ and write $y^x=\exp(x\ln y)=\exp\bigl((1+h)\ln(1+k)\bigr)$. 3. Use $\ln(1+k)=k-\tfrac{k^2}{2}+\cdots$ and expand exponent to second degree obtaining exponent $k+hk-\tfrac{k^2}{2}+\cdots$. 4. Then $y^x=\exp(\cdots)\approx 1+(k+hk)+\tfrac{1}{2}(k+hk)^2-\tfrac{k^2}{2}+\cdots$ and simplifying up to degree $2$ gives 5. $y^x\approx 1+(y-1)+(x-1)(y-1)$. 6. Hence at $(1,1)$ the value is $1$ and its reciprocal $(y^x)^{-1}$ at $(1,1)$ is also $1$. 1. Problem 22: Expand $xy^2+x^2y$ in powers of $(x-1)$ and $(y+3)$ up to second degree. 2. Let $h=x-1$ and $k=y+3$ so $x=1+h$ and $y=-3+k$ and expand powers keeping total degree $\le 2$. 3. Compute and collect terms to get 4. $xy^2+x^2y\approx 6+3(x-1)-5(y+3)-4(x-1)(y+3)+(y+3)^2-3(x-1)^2$. 1. Problem 23: Expand $\arctan(y/x)$ about $(1,1)$ up to third degree. 2. Put $h=x-1$ and $k=y-1$ and set $r=(1+k)/(1+h)=1+u$ with $u=(k-h)+(h^2-kh)+(-h^3+kh^2)+\cdots$. 3. Use the Taylor series about $1$: $\arctan(1+u)=\frac{\pi}{4}+\tfrac{1}{2}u-\tfrac{1}{4}u^2+\tfrac{1}{12}u^3+\cdots$ and expand keeping terms up to total degree $3$. 4. Collecting terms yields 5. $\arctan(y/x)\approx \dfrac{\pi}{4}+\tfrac{1}{2}(k-h)+\tfrac{1}{4}h^2-\tfrac{1}{4}k^2+\tfrac{1}{12}k^3+\tfrac{1}{4}k^2h-\tfrac{1}{4}kh^2-\tfrac{1}{12}h^3$. 6. Replace $h=x-1$ and $k=y-1$ to express the result in $x,y$. 1. Problem 25: Expand $e^x\log(1+y)$ in powers of $x,y$ up to total degree $3$ about $(0,0)$. 2. Use $e^x=1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\cdots$ and $\log(1+y)=y-\tfrac{y^2}{2}+\tfrac{y^3}{3}+\cdots$ and multiply keeping terms of degree $\le 3$. 3. Collecting terms gives 4. $e^x\log(1+y)\approx y+xy-\tfrac{y^2}{2}+\tfrac{y^3}{3}+\tfrac{x^2y}{2}-\tfrac{xy^2}{2}$.