Question

5. [0/4 Points] DETAILS PREVIOUS ANSWERS SCALCCC4 11.6.031. Suppose that over a certain region of space the electrical potential V is given by the following equation. V(x, y, z) = 8x2 - 5xy + xyz (a) Find the rate of change of the potential at P(4, 4, 6) in the direction of the vector v = i + j - k. (b) In which direction does V change most rapidly at P? <68, (c) What is the maximum rate of change at P?

Answer 1

Q5. Here given potential function is V (x, y, z) = 8x² - 5xy + xyz . Now WV (4.v.=)= ( (3x – 5xy + x3), n (8x2–5xy + xy2), {8x’ – 5xy + xyz] 197} = (16x – 5y+ yz, -5x + xz, xy). Then at P(4,4,6) we get, WV (4,4,6)= (16.4-5.4 +4:6, -5.4+4.6,4.4) = (64 – 20+ 24, -20+24,16) = (68,4,16) 1 1 Now the unit vector along the vector i+j-k is (i+j – k) ||i+j -k | (i+j - k) +1+(-1) + 1 (i+j-k) 13' 13' 13 1 Then rate of change of V at P(4,4,6) along i+j-k is VV. (i+j-k). i+j-k| 1 1 68 4 16 13' 15' 13 13 13 13 = + 68+4-16 V3 Now VV. (i+j – k)= (68,4,16). ||: + j– k|l 68+4-16 56 5613 √3 13 3 5613 a) Then rate of change of V at P(4,4,6) along i + j - k is 3 b) Then the maximum rate of change of V at P(4,4,6) is in the dire +V = (68,4,16) c) And the maximum rate of change of V at P(4,4,6) is +||VV|| = ||(64,4,16)||= V64² +42 +162 = (4x16) +42 +(4x4)? = V4(16² +1+42 ) = 42162 +1+4= 47256+1+16 = 4 273 Q3. Here given r(t) = 8V1 i+y? j-4tk,t=1 1 Now then P (0) - test i teºj-ark} = open kva) [ 1973) - 13. Ja k = 8x i +2t j - 4k 2Vt 4. >r'(t) i +2t j - 4k Vi 4 Then at t= 1 we get, r'(1) = i+2x1j- 4k = 4 i + 2 j - 4k. VI Then ||-' (1)|| = |(4,2,-4)= V42+2 +(-4)² = v16+4+16 = 136 = 6 4 2. 4 2 Then at t = 1 we get the unit tangent is T (t = 1) = ir' (t = 1) |:'(t=1) (4,2,-4) 2 1 33 3 =>T(1=1) = () *(]);+(-3) k