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A straight line going through $\boldsymbol{x}^{\circ}=\left(x_{1}^{\circ}, x_{2}...
Jan 29, 2024
A straight line going through x∘=(x1∘,x2∘)\boldsymbol{x}^{\circ}=\left(x_{1}^{\circ}, x_{2}^{\circ}\right) in the direction x=(x1,x2)≠0\boldsymbol{x}=\left(x_{1}, x_{2}\right) \neq 0 is the set of points x∘+tx=(x1∘+tx1,x2∘+tx2)\boldsymbol{x}^{\circ}+t \boldsymbol{x}=\left(x_{1}^{\circ}+t x_{1}, x_{2}^{\circ}+t x_{2}\right). A function f:R2→Rf: \mathbb{R}^{2} \rightarrow \mathbb{R} stays below its tangent plane if and only if it stays below its tangent plane on all straight lines going through all all x∘\boldsymbol{x}^{\circ} in all directions x≠0\boldsymbol{x} \neq 0. 2dim⁡2 \operatorname{dim} A. Suppose that the function f:R2→Rf: \mathbb{R}^{2} \rightarrow \mathbb{R} has continuous second derivatives (so that Young's theorem applies). For any x∘\boldsymbol{x}^{\circ} and x≠0\boldsymbol{x} \neq 0, let g(t)=f(x∘+tx)=g(t)=f\left(\boldsymbol{x}^{\circ}+t \boldsymbol{x}\right)= f(x1∘+tx1,x2∘+tx2)f\left(x_{1}^{\circ}+t x_{1}, x_{2}^{\circ}+t x_{2}\right). 1. Show that g′(0)=f1(x∘)x1+f2(x∘)x2=Df(x∘)⋅xg^{\prime}(0)=f_{1}\left(\boldsymbol{x}^{\circ}\right) x_{1}+f_{2}\left(\boldsymbol{x}^{\circ}\right) x_{2}=D f\left(\boldsymbol{x}^{\circ}\right) \cdot \boldsymbol{x}. 2. Show that g′′(0)=f11(x∘)x12+2f12(x∘)x1x2+f22(x∘)x22g^{\prime \prime}(0)=f_{11}\left(\boldsymbol{x}^{\circ}\right) x_{1}^{2}+2 f_{12}\left(\boldsymbol{x}^{\circ}\right) x_{1} x_{2}+f_{22}\left(\boldsymbol{x}^{\circ}\right) x_{2}^{2}.
Solution by Steps
step 1
To find g′(0) g'(0) , we differentiate f(x1∘+tx1,x2∘+tx2) f(x_1^{\circ} + t x_1, x_2^{\circ} + t x_2) with respect to t t and evaluate at t=0 t = 0
step 2
Using the chain rule, the derivative of f f with respect to t t is f1(x1∘+tx1,x2∘+tx2)⋅x1+f2(x1∘+tx1,x2∘+tx2)⋅x2 f_1(x_1^{\circ} + t x_1, x_2^{\circ} + t x_2) \cdot x_1 + f_2(x_1^{\circ} + t x_1, x_2^{\circ} + t x_2) \cdot x_2
step 3
Evaluating the derivative at t=0 t = 0 , we get g′(0)=f1(x1∘,x2∘)⋅x1+f2(x1∘,x2∘)⋅x2 g'(0) = f_1(x_1^{\circ}, x_2^{\circ}) \cdot x_1 + f_2(x_1^{\circ}, x_2^{\circ}) \cdot x_2
step 4
This can be written as a dot product: g′(0)=Df(x∘)⋅x g'(0) = Df(x^{\circ}) \cdot x , where Df(x∘) Df(x^{\circ}) is the gradient of f f at x∘ x^{\circ}
Answer
g′(0)=f1(x∘)x1+f2(x∘)x2=Df(x∘)⋅x g'(0) = f_1(x^{\circ}) x_1 + f_2(x^{\circ}) x_2 = Df(x^{\circ}) \cdot x
Key Concept
Differentiation along a path and evaluation at a point
Explanation
The derivative of a function along a path is found using the chain rule and evaluating at a specific point gives the directional derivative in the direction of the path.
Solution by Steps
step 1
To find g′′(0) g''(0) , we differentiate g′(t) g'(t) with respect to t t and evaluate at t=0 t = 0
step 2
The second derivative of f f with respect to t t is f11(x1∘+tx1,x2∘+tx2)⋅x12+2f12(x1∘+tx1,x2∘+tx2)⋅x1x2+f22(x1∘+tx1,x2∘+tx2)⋅x22 f_{11}(x_1^{\circ} + t x_1, x_2^{\circ} + t x_2) \cdot x_1^2 + 2 f_{12}(x_1^{\circ} + t x_1, x_2^{\circ} + t x_2) \cdot x_1 x_2 + f_{22}(x_1^{\circ} + t x_1, x_2^{\circ} + t x_2) \cdot x_2^2
step 3
Evaluating the second derivative at t=0 t = 0 , we get g′′(0)=f11(x1∘,x2∘)⋅x12+2f12(x1∘,x2∘)⋅x1x2+f22(x1∘,x2∘)⋅x22 g''(0) = f_{11}(x_1^{\circ}, x_2^{\circ}) \cdot x_1^2 + 2 f_{12}(x_1^{\circ}, x_2^{\circ}) \cdot x_1 x_2 + f_{22}(x_1^{\circ}, x_2^{\circ}) \cdot x_2^2
Answer
g′′(0)=f11(x∘)x12+2f12(x∘)x1x2+f22(x∘)x22 g''(0) = f_{11}(x^{\circ}) x_1^2 + 2 f_{12}(x^{\circ}) x_1 x_2 + f_{22}(x^{\circ}) x_2^2
Key Concept
Second derivative along a path
Explanation
The second derivative of a function along a path gives the curvature of the function in the direction of the path and is related to the Hessian matrix of second partial derivatives.
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