(20 pts) Consider the follow intensity values for an image...

(20 pts) Consider the follow intensity values for an image I: \begin{bmatrix} 4 & 4 & 4 & 2 & 2 \\ 4 & 4 & 4 & 2 & 2 \\ 4 & 4 & 4 & 1 & 1 \\ 2 & 2 & 2 & 1 & 1 \\ 2 & 2 & 2 & 1 & 1 \end{bmatrix} I = and answer the following equations. Enter you answer rounded to two decimal places. (a) (10 pts) Compute the gradient value $G = (\partial_x I, \partial_y I)$ associated with the bold underlined number in the image I. (b) (10 pts) Consider the following images patches below A,B, and C, and answer the following. A B C i. Consider the following plot of the gradient vector components $\partial_x I$ and $\partial_y I$. Which of the above image patches corresponds to the shown gradient plot (A, B, or C)? Iy Ix ii. Consider the following plot of the gradient vector components $\partial_x I$ and $\partial_y I$. Which of the above image patches corresponds to the shown gradient plot (A, B, or C)? Iy Ix

Transcript

00:01 Solve our second question of the given question where f of x comma y is given as x e to the power of i minus n x ren x so here the point p which is given as minus 3 and 0.

00:24 So a value for delta f will be equal to f n comma f y and therefore f of x x of x.

00:33 Will be equal to ey minus 1 upon x comma f of 5 is equal to ey x so therefore a value for delta f will be equal to e to the power of 5 minus 1 upon x comma e to the power of y x and then after we can say delta f at minus 3 comma will be e to the power of 0 minus minus 1 by minus 3 comma 8 to the power of 0 minus 3.

01:20 So this will be calculated as delta f minus 3 comma 0.

01:27 Sorry, this is not the minus sign over here and this will be equal to 1 by 3 comma minus 3.

01:35 So here we have our solution for first question which is given over here.

01:40 Answer for question number 2.

01:43 Now let's simplify our next question which is version number 4.

01:47 So in solution of question number 4 as given p is equal to x0 comma y not which is the particular or fixed point in the plane and here q which is x comma y is the variable variable point in the plane so to solve this question first of all we will make a space over here for the diagram and here we have two points in this is p whose coordinates as x0 comma y not and then this is q x comma y and p and p is x comma y and p is x comma y over here so here at the first part of the question, we find a distance from our distance formula of coordinate symmetry which is f of x comma y as for the formula x2 minus x1 whole square so x minus x not whole is squared added by y2 minus y1 whole square so this will be as y minus y not whole square and we observe that for any fixed point which is a non -negative depending non negative value of p x comma y because distance is here and the curve represent family of circles where if this is a family of circle then we have to find the radius and radius will be taken as f of x comma y so that is x minus x0 whole squared added by y is y0 whole square which is equal to f x comma y whole square so this can be written as x squared plus y square is equal to radius square r square so as for the observation we can say the level of curve of the family circle centered at x not and y not and having radius r is equal to f x.

04:41 Comma y so the center will be x not y not radius we have taken as f x comma y and where the value is greater than or equal to 0 and from this observation we can find our first solution which is f x comma y is equal to 0 this implies us as x minus x not whole square plus y minus y not whole sphere plus y minus y not whole sphere is equal to 0.

05:18 Which is a point on circle which is a point on circle...

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