Lesson 35 Exercise Review

Overall Assessment

The work demonstrates readiness for Lesson 36. The central ideas are secure: the gradient assembles coordinate sensitivities, directional derivatives are dot products, the gradient represents the differential in Euclidean coordinates, and gradient descent uses the negative gradient.

Problems 1-3, 6, and 8-11 are correct apart from local arithmetic or implementation cleanup. Problems 4, 5, and 7 need corrections, but the errors are sign-copying and geometric-expression issues rather than prerequisite gaps.

Problem-by-Problem Review

1. Correct overall. A partial derivative is one coordinate sensitivity, while the gradient collects all coordinate sensitivities. The surface-slice picture is useful. One geometric refinement: the gradient lives in the input plane and is normal to a contour line there; it is not a vector lying in the tangent plane of the graph itself.

2. Correct:

   $$\nabla f(x,y)= \begin{bmatrix} 6x-2y\\ -2x+2y \end{bmatrix}, \qquad \nabla f(1,2)= \begin{bmatrix} 2\\2 \end{bmatrix}.$$

3. The norm and stated unit direction are correct:

   $$\lVert\nabla f(1,2)\rVert_2=2\sqrt2, \qquad \frac{\nabla f(1,2)}{\lVert\nabla f(1,2)\rVert_2} =\frac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix} =\frac{\sqrt2}{2} \begin{bmatrix}1\\1\end{bmatrix}.$$

   The extra rationalization line at the bottom simplifies $1/(2\sqrt2)$ rather than the full vector expression and should be discarded.

4. All three direction vectors were correctly verified as unit vectors, but the gradient’s second component was copied as $+4$ instead of the exercise’s $-4$. Using

   $$\nabla f= \begin{bmatrix}3\\-4\end{bmatrix}$$

   gives:

   $$D_u f=3, \qquad D_v f=-\frac75, \qquad D_w f=-5.$$

5. The conclusion needs correction. If $t$ is tangent to the contour $f(x,y)=c$, moving along $t$ keeps $f$ constant to first order. Therefore:

   $$D_t f=\nabla f(x)^Tt=0.$$

   A zero dot product means the gradient is perpendicular to the contour tangent. The gradient is not tangent to the contour.

6. Correct. Cauchy-Schwarz gives the upper bound for every unit vector:

   $$\nabla f(x)^Tu \le \lVert\nabla f(x)\rVert_2\lVert u\rVert_2 =\lVert\nabla f(x)\rVert_2.$$

   Equality occurs when $u$ points in the gradient direction. The absolute-value form also bounds the magnitude of the directional derivative.

7. The setup identifies the right first-order formula and displacement, but $f$ was evaluated with the wrong sign and the gradient was not evaluated at $(1,-1)$. For

   $$f(x,y)=x^2+2y^2, \qquad \nabla f(x,y)= \begin{bmatrix}2x\\4y\end{bmatrix},$$

   we have:

   $$f(1,-1)=3, \qquad \nabla f(1,-1)= \begin{bmatrix}2\\-4\end{bmatrix}, \qquad \Delta x= \begin{bmatrix}0.02\\0.03\end{bmatrix}.$$

   Hence:

   $$f(1.02,-0.97)\approx3+2(0.02)-4(0.03)=2.92.$$

   The exact value is:

   $$1.02^2+2(0.97^2)=2.9222.$$

8. Correct. The displacement is $[-0.2,0.1,-0.2]^T$, and the predicted first-order loss change is $-0.9$.

9. Correct. A zero gradient supplies no first-order classification; the Hessian supplies the relevant second-order curvature information.

10. The analytic gradient is correct:

    $$\nabla f(1,-2)= \begin{bmatrix}0\\-11\end{bmatrix}.$$

    The centered-difference structure is also correct. The TypeScript needs small syntax fixes: close the function signature with ): number[], use one function name consistently, and return or assert the componentwise tolerance check.

11. Correct. The differential $df_x$ is the linear map that consumes an input displacement, while the gradient is the Euclidean vector representing that map through

    $$df_x(\Delta x)=\nabla f(x)^T\Delta x.$$

Readiness

Ready for Jacobians. Lesson 36 should reinforce three refinements:

• preserve signs when copying vectors and functions;
• derive contour orthogonality from a zero directional derivative;
• evaluate both the base function and gradient at the stated point before applying a first-order approximation.