Engineering Mathematics Problems focuses on “Using Properties of Divergence and Curl”.
1. Find the divergence of this given vector (vec{F}=x^3 yvec{i}+3xy^2 zvec{j}+3zxvec{k}).
a) 3x2 y+6xyz+x
b) 2x2 y+6xyz+3x
c) 3x2 y+3xyz+3x
d) 3x2 y+6xyz+3x
View Answer
Answer: d
Explanation: We know that divergence of a vector is given by
(bigtriangledown.vec{F}=frac{partial(x^3 y)}{partial x}+frac{partial(3xy^2 z)}{partial y}+frac{partial(3xz)}{partial z})
(bigtriangledown.vec{F}=3x^2 y+6xyz+3x).
2. Find the divergence of this given vector (vec{r}=12x^6 y^6 vec{i}+3x^3 y^3 zvec{j}+3x^2 yz^2 vec{k}).
a) (12x^5 y^6+2x^3 yz+6x^2 yz)
b) (72x^5 y^6+2x^3 yz+3x^2 yz)
c) (72x^5 y^6+2x^3 yz+6x^2 yz)
d) (6x^5 y^6+2x^3 yz+6x^2 yz)
View Answer
Answer: c
Explanation: We know that divergence of a vector is given by
(bigtriangledown.vec{r}=frac{partial(12x^6 y^6)}{partial x}+frac{partial(3x^3 y^3 z)}{partial y}+frac{partial(3x^2 yz^2)}{partial z})
(bigtriangledown.vec{r}=12×6x^5 y^6+x^3×2y×z+3x^2×y×2z)
(bigtriangledown.vec{r}=72x^5 y^6+2x^3 yz+6x^2 yz).
3. Find the curl for (vec{r}=x^2 yzvec{i}+(3x+2y)zvec{j}+21z^2 xvec{k}).
a) (vec{i}(3x+2y)-vec{j}(11z^2-x^2 y)+vec{k}(3z-x^2 z))
b) (vec{i}(x+2y)-vec{j}(21z^2-x^2 y)+vec{k}(3z-x^2 z))
c) (-vec{i}(3x+2y)-vec{j}(21z^2-x^2 y)+vec{k}(3z-x^2 z))
d) (vec{i}(3x+2y)-vec{j}(21z^2-x^2 y)+vec{k}(3z-x^2 z))
View Answer
Answer: c
Explanation: We know that the curl for any vector quantity is given by
(bigtriangledown.vec{r}=begin{bmatrix}vec{i}&vec{j}&vec{k}\frac{partial}{partial x}&frac{partial}{partial y}&frac{partial}{partial z}\x^2 yz&(3x+2y)z&21z^2 xend{bmatrix})
(bigtriangledown.vec{r}=vec{i}left (frac{partial(21z^2 x)}{partial y}-frac{partial((3x+2y)z)}{partial z}right)-vec{j}left (frac{partial(21z^2 x)}{partial x}-frac{partial(x^2 yz)}{partial z}right))
(+vec{k}left (frac{partial((3x+2y)z)}{partial x}-frac{partial(x^2 yz)}{partial y}right))
(bigtriangledown.vec{r}=-vec{i}(3x+2y)-vec{j}(21z^2-x^2 y)+vec{k}(3z-x^2 z)).
4. Find the curl for ((vec{r})=y^2 z^3 vec{i}+x^2 z^2 vec{j}+(x-2y)vec{k}).
a) (-2vec{i}(1+x^2 z)-vec{j}(1-3y^2 z^2)+vec{2k}(xz^2-yz^3))
b) (-2vec{i}(1+x^2 z)-vec{j}(1-3y^2 z^2)+vec{k}(xz^2-yz^3))
c) (-2vec{i}(1+x^2 z)-vec{j}(1-32z^2)+vec{2k}(xz^2-yz^3))
d) (vec{i}(1+x^2 z)-vec{j}(1-3y^2 z^2)+vec{2k}(xz^2-yz^3))
View Answer
Answer: a
Explanation: We know that the curl for any vector quantity is given by
(bigtriangledown.vec{r}=begin{bmatrix}vec{i}&vec{j}&vec{k}\partial/partial x&partial/partial y&partial/partial z\y^2 z^3&x^2 z^2&(x-2y)end{bmatrix})
(bigtriangledown.vec{r}=vec{i}left (frac{partial(x-2y)}{partial y}-frac{partial(x^2 z^2)}{/partial z}right )-vec{j}left (frac{partial(x-2y)}{partial x}-frac{partial(y^2 z^3)}{partial z}right )+vec{k}left(frac{partial(x^2 z^2)}{partial x}-frac{partial(y^2 z^3)}{partial y}right ))
(bigtriangledown.vec{r}=-2vec{i}(1+x^2 z)-vec{j}(1-3y^2 z^2)+vec{2k}(xz^2-yz^3)).
5. What is the divergence and curl of the vector (vec{F}=x^2 yvec{i}+(3x+y) vec{j}+y^3 zvec{k}).
a) (y^3+2xy+1,vec{i}(3y^2 z)+vec{j}(3-x^2))
b) (y^3+2xy+1,vec{i}(3y^2 z)+vec{k}(3-x^2))
c) (3y^3+2xy+1,vec{i}(3y^2 z)+vec{k}(3-x^2))
d) (y^3+xy+1,vec{i}(3y^2 z)+vec{k}(3-x^2))
View Answer
Answer: b
Explanation: We know that divergence of a vector is given by
(bigtriangledown.vec{F}=frac{partial(x^2 y)}{partial x}+frac{partial(3x+y)}{partial y}+frac{partial(y^3 z)}{partial z})
(bigtriangledown.vec{F}=y^3+2xy+1)
We know that the curl for any vector quantity is given by
(bigtriangledown.vec{r}=begin{bmatrix}vec{i}&vec{j}&vec{k}\frac{partial}{partial x}&frac{partial}{partial y}&frac{partial}{partial z}\x^2 y&3x+y&(y^3 z)end{bmatrix})
(bigtriangledown.vec{r}=vec{i}left (frac{partial(y^3 z)}{partial y}-frac{partial(3x+y)}{partial z}right )-vec{j}left (frac{partial(y^3 z)}{partial x}-frac{partial(x^2 y)}{partial z}right )+vec{k}left (frac{partial(3x+y)}{partial x}-frac{partial(x^2 y)}{partial y}right ))
(bigtriangledown.vec{r}=vec{i}(3y^2 z)+vec{k}(3-x^2)).
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