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Q.4. The directional derivative of ø= 4x2 – 3rºyz? at (2. -1, 2) along Z-axis is
(a) 144
(b) 133
(c) 331
(d) 441
Answers
Answer:
Example 1
Let f(x,y)=x2y. (a) Find ∇f(3,2). (b) Find the derivative of f in the direction of (1,2) at the point (3,2).
Solution: (a) The gradient is just the vector of partial derivatives. The partial derivatives of f at the point (x,y)=(3,2) are:
∂f∂x(x,y)∂f∂x(3,2)=2xy=12∂f∂y(x,y)∂f∂y(3,2)=x2=9
Therefore, the gradient is
∇f(3,2)=12i+9j=(12,9).
(b) Let u=u1i+u2j be a unit vector. The directional derivative at (3,2) in the direction of u is
Duf(3,2)=∇f(3,2)⋅u=(12i+9j)⋅(u1i+u2j)=12u1+9u2.(1)
To find the directional derivative in the direction of the vector (1,2), we need to find a unit vector in the direction of the vector (1,2). We simply divide by the magnitude of (1,2).
u=(1,2)∥(1,2)∥=(1,2)12+22−−−−−−√=(1,2)5√=(1/5√,2/5√).
Plugging this expression for u=(u1,u2) into equation (1) for the directional derivative, and we find that the directional derivative at the point (3,2) in the direction of (1,2) is
Duf(3,2)=12u1+9u2=125√+185√=305√.
Example 2
For the f of Example 1, find the directional derivative of f at the point (3,2) in the direction of (2,1).
Solution: The unit vector in the direction of (2,1) is
u=(2,1)5√=(2/5√,1/5√).
Since we are still at the point (3,2), equation (1) is still valid. We plug in our new u to obtain
Duf(3,2)=12u1+9u2=245√+95√=335√
Example 3
For the f of Example 1 at the point (3,2), (a) in which direction is the directional derivative maximal, (b) what is the directional derivative in that direction?
Solution: (a) The gradient points in the direction of the maximal directional derivative. The directional derivative is maximal in the direction of (12,9). (A unit vector in that direction is u=(12,9)/122+92−−−−−−−√=(4/5,3/5).)
(b) The magnitude of the gradient is this maximal directional derivative, which is ∥(12,9)∥=122+92−−−−−−−√=15. Hence the directional derivative at the point (3,2) in the direction of (12,9) is 15.
We could double-check by calculating the result using equation (1) and the unit vector u=(4/5,3/5). Then we find that
Duf(3,2)=12u1+9u2=485+275=755=15,
which agrees with our result.
Example 4
For the f of Example 1 at the point (3,2), (a) what is the directional derivative in the direction (-3,4) (which is perpendicular to ∇f(3,2)), and (b) what is the directional derivative in the direction (-4,-3) (which is opposite of the direction of ∇f(3,2))?
Solution: (a) The directional derivative must be zero. (b) The directional derivative must be −∥∇f(3,2)∥, which is −15. (You can verify these by calculating the results directly using equation (1).)
Example 5
Let f(x,y,z)=xyex2+z2−5. Calculate the gradient of f at the point (1,3,−2) and calculate the directional derivative Duf at the point (1,3,−2) in the direction of the vector v=(3,−1,4).
Solution: The gradient vector in three-dimensions is similar to the two-dimesional case. To calculate the gradient of f at the point (1,3,−2) we just need to calculate the three partial derivatives of f.
∇f(x,y,z)∇f(1,3,−2)=(∂f∂x,∂f∂y,∂f∂z)=((y+2x2y)ex2+z2−5,xex2+z2−5,2xyzex2+z2−5)=(3+2(1)23e0,1e0,2(1)(3)(−2)e0)=(9,1,−12)
Just as for the above two-dimensional examples, the directional derivative is Duf(x,y,z)=∇f(x,y,z)⋅u where u is a unit vector. To calculate u in the direction of v, we just need to divide by its magnitude. Since ∥v∥=32+(−1)2+42−−−−−−−−−−−−−√=26−−√,
u=v26−−√=(326−−√,−126−−√,426−−√)
and
Duf(1,3,−2)=∇f(1,3,−2)⋅u=(9,1,−12)⋅(326−−√,−126−−√,426−−√)=9⋅3−1−12⋅426−−√=−2226−−√.
Step-by-step explanation:
Answer:
Step-by-step explanation: