The temperature in the room is is given by $$T(x,y,z) = \frac12(4x^2 +y^2 + z^2)$$ At the time $t = 0$ a fly flies through the point $(\sqrt{3},1,2)$, along the curve of intersection between the two surfaces $$ z = x^2 - y^2$$ $$z^2 = x^2 + y^2$$ What change of temperature $$\frac{dT}{dt}_{t=0} $$ does the fly experience at $t = 0$ if its vertical velocity at this time is $5$?
Calling the surfaces $f$ and $g$ I go about finding the tangent vector of the curve of intersection: $\nabla f \times \nabla g = \begin{bmatrix}-10 \\ -6\sqrt{3} \\ 8\sqrt{3} \end{bmatrix}$ then i scale this to match the speed of the fly $(z'(0) = 5)$ and get $\begin{bmatrix} -\frac{25}{4\sqrt{3}} \\ -\frac{15}{4} \\ 5 \end{bmatrix}$ and finally find the rate of change which is given by $\nabla T \cdot \vec{v} $ where $\vec{v}$ is the scaled velocity vector. This gives me the answer $-\frac{75}{4}$ which appears to be wrong.
The two surfaces are level surfaces of $f(x,y,z)=x^2-y^2-z$ and $g(x,y,z)=x^2+y^2-z^2$, respectively. $\nabla f\times\nabla g = (4yz+2y,4xz-2x,8xy)$, which is equal to $(10,6\sqrt3,8\sqrt3)$ at the given point. It looks like you forgot the minus signs when constructing $f$ and $g$ and computed the gradients of $x^2-y^2+z$ and $x^2+y^2+z^2$ instead.