Heavy slope

The gradient of $f(x,y)=x^{-2}+y^3+3x$ is defined as the vector $\nabla f =(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})=(3-2x^{-3}, 3y^2)$ A gradient is just a derivative, and it determines the steepest tangent line as well as its slope. It takes the value $\nabla f (1,-2)=(1, 12)$ In general the slope of the graph of a 2-dim. real-valued function is given by $|\nabla f(x,y)|$ so the angle is given by $\arctan(\sqrt{1^2+12^2}) \approx \boxed{85.253°}$

A surface determined by the graph of a function $f$ can be written as the implicit surface equation $g(x,y,z) = 0$ , where $g(x,y,z)=f(x,y)-z$ . In turn, the gradient of $g$ at $(x,y,z)$ gives the components of the vector in the same direction as the vector normal to the plane which is tangent to the graph of $f$ at this same point. Thus, $\nabla g (x,y,z) = \left(-\frac{2}{x^3} + 3, 3y^2, -1\right)$ , $\nabla g (1,-2,f(1,-2)) = \left(1, 12, -1\right)$ , and the tangent plane $\pi: \ (x-1)+12(y+2)-(z+4) = 0$ . Setting to zero two out of the three variables in the plane equation, one finds out that the following points belong to $\pi$ : $(0, 0, 19)$ , $(0, -\frac{19}{12}, 0)$ and $(-19, 0, 0)$ . Hence, $(19, 0, 19) \propto (1,0,1)$ and $(0, \frac{19}{12}, 19) \propto (0, 1, 12)$ are direction vectors of this plane. Since the tangent line belongs to $\pi$ , its direction vector must be a linear combination of the direction vectors of the plane, i.e. $\lambda (1,0,1) + \mu (0, 1, 12) = (\lambda, \mu, \lambda + 12 \mu)$ . The angle between the tangent line and the $z=0$ plane must be maximized, which means that the angle between this vector and the normal to the $z=0$ plane must be minimized; in turn, this means that the cosine of the angle between this vector and the normal to the $z=0$ plane ( $cos \ \alpha$ ) must be maximized. The cosine of the given angle is simply:

$cos \ \alpha = \frac{\langle (\lambda, \mu, \lambda + 12 \mu), (0,0,1)\rangle}{\mid \mid (\lambda, \mu, \lambda + 12 \mu) \mid \mid \cdot \mid \mid (0,0,1)\mid \mid} = \frac{\lambda + 12 \mu}{\sqrt{\lambda^2 + \mu^2 + \left(\lambda+12 \mu\right)^2}}$

Taking the derivatives of this expression with respect to $\lambda$ and $\mu$ , equating to zero and solving the resulting system of equations, one finds out that $\lambda = \frac{\mu}{12}$ and that $\mu$ can be any real number (the other solution would give a null vector as the direction vector). Thus, the direction vector is simply $\mu(\frac{1}{12}, 1, \frac{145}{12}) \propto (1, 12, 145)$ . Therefore, $cos \ \alpha = 145 \cdot \left(1^2 + 12^2 + 145^2\right)^{-1/2} \approx 0.996569 \Rightarrow \alpha \approx 4.747^{\circ}$ and the sought-after angle is its complement, $85.253^{\circ}$ .

Since $z= f(x, y)$ , then $g(x, y, z) = f(x, y) - z = 0$ . The gradient of $g$ is

$\nabla g = \begin{pmatrix} \nabla f \\ -1 \end{pmatrix} = \begin{pmatrix} -2 x^{-3} + 3 \\ 3 y^2 \\ -1 \end{pmatrix}$

At $(1, -2)$ , we have

$\nabla g = \begin{pmatrix} 1 \\ 12 \\ -1 \end{pmatrix}$

The gradient $\nabla g$ is normal to the tangent plane to the surface $g = 0$ . The maximum angle between a line in this plane and the $z = 0$ plane is the acute angle between $\nabla g$ and $\mathbf{k} = (0, 0, 1)$ the unit normal to the $z =0$ plane. Hence, the required angle is

$\theta = \displaystyle \cos^{-1} | \dfrac{\nabla g \cdot \mathbf{k}}{ | \nabla g | } | = \cos^{-1} \dfrac{1}{\sqrt{1^2 + 12^2 + 1^2}} = \cos^{-1} \dfrac{1}{\sqrt{146}} = 85.25273^{\circ}$

The surface has normal vector $\left( \begin{array}{c} 3 - 2x^{-2} \\ 3y^2 \\ -1 \end{array}\right)$ and hence, when $x=1,y=-2$ , we have $z = f(1,-2) = -4$ and

$\mathbf{n} \; = \; \left(\begin{array}{c} 1 \\ 12 \\ -1 \end{array}\right)$ Thus the tangent plane to the surface at the point $P\; (1,-2,f(1,-2))$ has equation $x + 12y - z + 19 = 0$ , which meets the $z=0$ plane in the line $x + 12y + 19 = 0\,,\,z=0$ . The closest distance from this line to the point $(1,-2)$ is $d \; = \; \left| \frac{1 - 24 + 19}{\sqrt{1 + 12^2}}\right| \; = \; \frac{4}{\sqrt{145}}$ and hence the largest angle that a tangent line at $P$ makes with the $z=0$ plane is $\tan^{-1} \tfrac{4}{d} \; = \; \tan^{-1}\sqrt{145} \; = \; \boxed{85.25273773^\circ}$

At $(x,y)=(1,-2)$ , the gradient vector is found by taking the partial derivatives of f in x and y: $\nabla f(1,-2)=\left.\left(-\frac2{x^3}+3,\ 3y^2\right)\right|_{(1,-2)}=(1,12)$ In a direction $\mathbf u=(\cos\theta,\sin\theta)$ , the slope at this point is $\mathbf u\cdot\nabla f(1,-2)=\cos\theta+12\sin\theta$ We can rewrite this sum of trig functions as $\sqrt{1^2+12^2}\sin(\theta+\phi)=\sqrt{145}\sin(\theta+\phi)$ for some angle $\phi$ . This slope attains a maximum value of $\sqrt{145}$ , hence the maximum angle between the tangent line and the $z=0$ plane is $\arctan(\sqrt{145})\approx85.253^\circ$