Calculating the Directional Derivative of T at (1,1,2)

[twoflower]

$$T(x,y,z) = x^2 - y^2 + z^2 + xz^2$$

$$3x^2 - y^2 - z = 0$$

$$2x^2 + 2y^2 - z^2 = 0$$

$$\bigtriangledown T (x,y,z) = (2x + z^2, -2y, 2z(1+x))$$
$$\bigtriangledown T (1,1,2) = 2(3, -1, 4)$$
$$D_{\overrightarrow{v}}(1,1,2) = \bigtriangledown T(1,1,2).\overrightarrow{v} = 2(3v_1 - v_2 + 4v_3)$$
$$T_1 = (6x, -2y, -1)$$
$$T_2 = (4x, 4y, -2z)$$
$$T_1 = (6, -2, -1)$$
$$T_2 = (4, 4, -4)$$
$$\overrightarrow{n} = T_1\ \mbox{x}\ T_2 = (12, 20, 32) = 4(3,5,8)$$
$$||\overrightarrow{n}|| = 4\sqrt{9 + 25 + 64} = 28\sqrt{2}$$
$$\frac{2.7.4}{28\sqrt{2}} (3.3 - 5 + 4.8) = 33\sqrt{2} \frac{°C}{s}$$

 

[twoflower]

I don't know how to delete my thread, I was preparing the TeX graphics and in the meanwhile I already found the solution...please could someone delete it?

Thank you and sorry.