Take, for instance, problem e. From taking two partial derivatives in the \(x\) and \(y\) dimensions, we deduce that the partial derivative values are…
We could, therefore, treat these terms as two separate vectors lying at the \(x\) and \(y\) directions. That is, we know that the multidimentional "slope" of the function could be represented by a combination of vectors…
\begin{equation} \left\{\begin{pmatrix}10 \\ 0 \end{pmatrix},\begin{pmatrix}0 \\ 10\sqrt{3} \end{pmatrix} \right\} \end{equation}The "slope" created by the two slope values at a \(60^{\circ}\) angle is essential the sums of the two partial derivative vectors projected at \(60^{\circ}\). Hence, we have to project the two vectors' magnitudes to a shared \(60^{\circ}\) angle, and sum it up.
We first note that, to project the two orthogonal vectors to the same, shared "60-degrees" direction, we must project one vector to \(60^{\circ}\) and the other to \((90-60)^{\circ}\) to actually result in the projections' alignment.
Conventionally, we will project the x-direction vector to \(60^{\circ}\). and the y-direction vector to \((90-60)^{\circ}\), but the 60 degree direction that we aim to share is actual arbitrary.
To perform the actual magnitude projection, we perform the follows.
\begin{align} let\ \vec{X} &= \begin{pmatrix}10\\0\end{pmatrix} \\ \vec{Y} &= \begin{pmatrix}0\\10\sqrt{3}\end{pmatrix} \\ \vec{X_p} &= \|X\|cos(60^{\circ}) \\ &= 10 \times \frac{1}{2} \\ &= 5 \\ \vec{Y_p} &= \|Y\|cos((90-60)^{\circ}) \\ &= \|Y\|sin(60^{\circ}) \\ &= 10\sqrt{3}\times\frac{\sqrt{3}}{2} \\ &= \frac{30}{2} = 15 \end{align}Finally, the sum of slopes in that shared direction would therefore be \(20\).