Question

3. Figure showing level curves off (x,y), estimate the following derivatives. 1 2 3 4 5 6 a) f(3,1) b) f,(3,1) c) Draw the gradient vector off at the point (3,1). d) fi(3,1) where üri- j e) fu(3,1) where ü = 27 +j

Answer 1

& In the given figure which is showing level curven of fixso), we can easily find the curve fcx, y) = 1 passes through (3.1). To find fx (21) we need to find a very close point to (3,1) through which any level Curve is passing through. Here we find *(5,1) is that required point as the level curre fex,y)=2 parses through it. So, we can conclude that fe (3,1) est at z f 13th, 1) - £13») ho 1 2 +(3+1, 1)-f(3,1) [we consiten halas f(4,1) is known to us] 2 H4,1) –f(331), 35121 (Any) by Like the previous problem to find fy 13,1) we I already know f(3,1)=1, so, we need another point of the form $ (3,1+k) for which +13,14k) in known on in other words a level curre passes through it. From the given a figure we can easily see that if we consider k=a, then the point in 13,6) and the level curre parses through it is fexy)=3 . . f(3,1+x) – $13,1 , fon K=Q-1 K-6 z f(3,0) =f(3,1) = 3.01 2 $13 . 23 --2 Com) Now, we have to draw the gradient vector at the point (3,1). Before drawing we need to define it at first. 5$) of) to fali-2 13.) 21-23 oun estimation) So, fy(3,1) 2 f(3,1+*) – $13,1). (-1) BRUCE ( as ken

Here î and ŷ re nenit reators along positre direction of x and yoxes respectively. Now, we can have drawn î in the positive direction of a axis and - 2 ŷ in the negatire direction of of axis, the P13) resultant in î- z i on all is chawn by the help of paralleogram law .. of the defimation of $1434) - Of] So, for 33 )2 (-23). (8-5) [w=-3) ♡ 2 ♡ Uw In Ô z 5. izog and e lait bo Iz razto 7 es like the previous problem, fu (3,1) = 5$] .czî+1) (3,1) 12ptil -23). (zi+) N 2² +12 - 2 2 - 0 (Aus)