The height of mountain is given by $z = y^5-2x^2y+1$ at the coordinate $(x,y)$. Give the 3-D dimensional direction in which a marble begins to roll if it is related at $(-1,-1,2)$.
Progress: I have calculated the gradient of $z=f(x,y)$ at given point, but could not figure out how to evaluate direction of rolling marble. Any help would be kindly appreciated!
On
Hint. The gradient of $f(x,y)$ points in the direction of fastest increase of $f.$ Thus, one vector which satisfies your condition is $-\nabla f_P.$
On
Imagine a nonhorizontal plane $\tau: \ z=ax+by$. A marble left rolling at the origin $O\in\tau$ will roll along the fall line $f\subset \tau$. This $f$ is the steepest line through $O$ on $\tau$. You know that it can be characterized as follows: It is lying in $\tau$, and is orthogonal to the horizontal line in $\tau$.
In the situation at hand $\tau$ is the tangential plane $\tau_P$ to the mountain $M:\ z= y^5-2x^2y+1$ at the point $P:=(-1,-1,2)\in M$. The total orbit of the marble will be a "curved curve" on $M$ (and is difficult to compute), but the starting direction is the direction of $f$ in the foregoing paragraph.
The mountain $M$ is a level surface of the function $$g(x,y,z):=y^5-2x^2y+1-z\ .$$ The normal vector of $\tau_P$ is given by $\nabla g(P)=(-4,3,-1)$. We furthermore need a horizontal vector in $h\in \tau_P$. This vector has the form $h=(u,v,0)$ and should satisfy $\nabla g(P)\cdot h=0$. We therefore may take $h=(3,4,0)$. The direction of the fall line $f$ then is orthogonal to $\nabla g(P)$ and $h$, and therefore is proportional to $$\nabla g(P)\times h=(-4,3,-1)\times(3,4,0)=(4,-3,-25)\ .$$ The parallel unit vector with negative $z$-component is then given by $${1\over\sqrt{2}}\left({4\over5},\ -{3\over5},\ -1\right)\ ,$$ and this is the initial direction of your marble.
Let $g(x,y,z)=y^5-2x^2y-z$; then your mountain is the level surface$$M=\{(x,y,z)\mid g(x,y,z)=-1\}.$$Let $P=(-1,-1,2)$. Then $\nabla g(P)=(-4, 3, -1)$. So, the tangent plane $p$ of $M$ at $P$ is the plane orthogonal to $(-4,3,-1)$ passing through $P$. The marble will start moving along a vector $\overrightarrow{PQ}$ with $Q\in p$. You know that $\overrightarrow{PQ}$ is orthogonal to $(-4,3,-1)$. Another vector which is orthogonal to $(-4,3,-1)$ is $(3,4,0)$ which is a horizontal vector. So, the marble will start moving along a direction which is orthogonal both to $(-4,3,-1)$ and to $(3,4,0)$. So, simply take their vector product, which is $(4,-3,-25)$. Then, take$$Q=P+(4,-3,-25)=(3,-4,-23);$$the marble will start moving along the direction given by $\overrightarrow{PQ}$.
You can see this in the picture below: the blue surface is $M$, the green line is the line passing through $P$ with the direction given by the gradient of $g$ at $P$, the yellow plane is the tangent plane of $M$ at $P$ and the vector shows the direction along which the marble will start moving (I used a point closer to $P$ than the one that I got in my answer, so that everything would fit confortably in one picture).