Mathematics-I (3110014)

BE | Semester-1   Winter-2019 | 17-01-2020

Q5) (a)

Find the directional derivatives of f=xy2+yz2 at the point 2,-1,1, in the direction of i+2j+2k.

We have, f=xy2+yz2 and u¯=i+2j+2k=(1,2,2).

fx=y2             fx(2,-1,1)=(-1)2=1

fy=2xy+z2   fy(2,-1,1)=2(2)(-1)+(1)2=-4+1=-3

fx=2yz          fx(2,-1,1)=2(-1)(1)=-2

Now,

f=fx i^+fy j^+fz k^

(f)(2,-1,1)=fx(2,-1,1)i^+fy(2,-1,1)j+fz(2,-1,1)k^

(f)(2,-1,1)=1 i^+-3 j+-2 k^

(f)(2,-1,1)=1, -3, -2

Now, u^=uu

Now, u^=(1,2,2)12+22+22

Now, u^=(1,2,2)9

Now, u^=(1,2,2)3

For Directional Derivative,

(f)(2,-1,1)  u^=(1,-3,-2)  (1,2,2)3

(f)(2,-1,1)  u^=(1)(1)+(-3)(2)+(-2)(2)3

(f)(2,-1,1)  u^=1-6-43

(f)(2,-1,1)  u^=-93

(f)(2,-1,1)  u^=-3

So, Directional Derivative is 3.