I need to find a function $g(x,y,t)$ that it's polynomial with degree <=2, Such that the Directional derivative of $f(x,y)=\int_{-7}^{8} g(x,y,t) dt$ in point $(2,7)$ with direction $u=\frac {-4}{97^{0.5}}i+\frac{9}{97^{0.5}}j$ is equal to 4.
How may I solve this kind of questions, are tried more than 50 functions and non worked, Should I really depend on luck and tries?
Note: The answer is expected to look like: $K_1x^2+K_2y^2+K_3t^2+K_4xy+K_5xt+K_6yt+K_7x+K_8y+K_9t+K_{10}$
That shouldn't be this hard. Let $$g(x,y,t)=K_1x^2+K_2y^2+K_3t^2+K_4xy+K_5xt+K_6yt+K_7x+K_8y+K_9t+K_{10}$$ $$G(x,y,t)=\int g(x,y,t)\,\mathrm{d}t=\\ K_1x^2t+K_2y^2t+\frac13K_3t^3+K_4xyt+\frac12K_5xt^2+\frac12K_6yt^2+\\K_7xt+K_8yt+\frac12K_9t^2+K_{10}t+C$$ $$f(x,y)=G(x,y,8)-G(x,y,-7)=\\ 15K_1x^2+15K_2y^2+285K_3+15K_4xy+\frac{15}{2}K_5x+\\ \frac{15}{2}K_6y+15K_7x+15K_8y+\frac{15}{2}K_9+15K_{10}$$ $$\frac{\partial f}{\partial x}=30K_1x+15K_4y+\frac{15}{2}K_5+15K_7$$ $$\frac{\partial f}{\partial y}=30K_2y+15K_4x+\frac{15}{2}K_6+15K_8$$ $$\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)_{(x,y)=(2,7)}=\\ (60K_1+105K_4+\frac{15}{2}K_5+15K_7, 210K_2+30K_4+\frac{15}{2}K_6+15K_8)$$ $$\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)_{(x,y)=(2,7)}.u=\\ \frac{1}{\sqrt{97}}\left( -240K_1-420K_4-30K_5-60K_7 +1890K_2+270K_4+\frac{135}{2}K_6+135K_8 \right)$$ So if we let all but $K_4$ equal $0$ we'll have $$-\frac{150}{\sqrt{97}}K_4=4\hbox{ thus }K_4=-\frac{2\sqrt{97}}{75}.$$