\[f(x,y,z) = y^2e^{xyz}, P(0,1,-1), u = <\frac{3}{13},\frac{4}{13},\frac{12}{13}>\]
This question aims to find the rate of change or gradient and projections of vector spaces onto a given vector.
Gradient of a vector can be found using following formula:
\[\nabla f(x,y,z) = \bigg ( \frac{\partial f}{\partial x} (x,y,z),\frac{\partial f}{\partial y} (x,y,z),\frac{\partial f}{\partial z} (x,y,z) \bigg )\]
Projection of a vector space can be found using dot product formula:
\[D_uf(x,y,z) = \nabla f(x,y,z) \cdot u\]
To solve the question, we will use the following steps:
- Find partial derivatives.
- Find the gradient.
- Find the projection of gradient in the direction of the vector $u$.
Expert Answer
Calculating partial derivative w.r.t $x$:
\[\frac{\partial f}{\partial x} (x,y,z) = \frac{\partial}{\partial x}\bigg ( y^2e^{xyz} \bigg )= y^2e^{xyz}(yz) = y^3ze^{xyz}\]
Calculating partial derivative w.r.t $y$:
\[\frac{\partial f}{\partial y} (x,y,z) = \frac{\partial}{\partial y}\bigg ( y^2e^{xyz} \bigg ) \]
\[\frac{\partial f}{\partial y} (x,y,z) = \frac{\partial}{\partial y} (y^2) e^{xyz} + y^2\frac{\partial}{\partial y} (e^{xyz}) \]
\[\frac{\partial f}{\partial y}(x,y,z) = 2y^2e^{xyz}+y^2e^{xyz}(xz) \]
\[\frac{\partial f}{\partial y}(x,y,z) = 2y^2e^{xyz} +xy^2ze^{xyz} \]
Calculating partial derivative w.r.t $z$:
\[\frac{\partial f}{\partial z} (x,y,z) = \frac{\partial}{\partial z}\bigg ( y^2e^{xyz} \bigg )= y^2e^{xyz}(xy) = xy^3e^{xyz}\]
Evaluating all partial derivatives at the given point $P$,
\[\frac{\partial f}{\partial x} (0,1,-1) = (1)^3(-1)e^{(0)(1)(-1)} = -1\]
\[\frac{\partial f}{\partial y} (0,1,-1) = 2(1)^2e^{(0)(1)(-1)}+(0)(1)^2(-1)e^{(0)(1)(-1)} = 2\]
\[\frac{\partial f}{\partial z} (0,1,-1) = (0)(1)^3e^{(0)(1)(-1)} = 0\]
Calculating the gradient of $f$ at point $P$:
\[\nabla f(x,y,z) = \bigg ( \frac{\partial f}{\partial x} (x,y,z),\frac{\partial f}{\partial y} (x,y,z),\frac{\partial f}{\partial z} (x,y,z) \bigg )\]
\[\nabla f (0,1,-1) = \bigg ( \frac{\partial f}{\partial x} (0,1,-1),\frac{\partial f}{\partial y} (0,1,-1),\frac{\partial f}{\partial z} (0,1,-1) \bigg )\]
\[\nabla f (0,1,-1) = < -1 , 2 , 0 >\]
Calculating the rate of change in the direction of $u$:
\[D_uf(x,y,z) = \nabla f(x,y,z) \cdot u\]
\[D_uf (0,1,-1) = \nabla f (0,1,-1) \cdot <\frac{3}{13},\frac{4}{13},\frac{12}{13}>\]
\[D_uf (0,1,-1) = <-1 , 2 , 0 > \cdot <\frac{3}{13},\frac{4}{13},\frac{12}{13}>\]
\[D_uf (0,1,-1) = -1(\frac{3}{13}) + 2(\frac{4}{13}) + 0(\frac{12}{13}) \]
\[D_uf (0,1,-1) = \frac{-1(3) + 2(4) + 0(12)}{13} \]
\[D_uf (0,1,-1) = \frac{-3 + 8 + 0}{13} = \frac{5}{13} \]
Numerical Answer
The rate of change is calculated to be:
\[ D_uf (0,1,-1) = \frac{5}{13} \]
Example
We have the following vectors and we need to calculate the rate of change.
\[ f(x,y,z) = y^2e^{xyz}, P(0,1,-1), u = <\frac{1}{33},\frac{5}{33},\frac{7}{33}> \]
Here, partial derivatives and the gradient values remain same, So:
\[ \frac{\partial f}{\partial x} (x,y,z) = y^3ze^{xyz} \]
\[ \frac{\partial f}{\partial y} (x,y,z) = 2y^2e^{xyz}+xy^2ze^{xyz} \]
\[ \frac{\partial f}{\partial z} (x,y,z) = xy^3e^{xyz} \]
\[ \frac{\partial f}{\partial x} (0,1,-1) = -1 \]
\[ \frac{\partial f}{\partial y} (0,1,-1) = 2\]
\[ \frac{\partial f}{\partial z} (0,1,-1) = 0\]
\[ \nabla f (0,1,-1) = < -1 , 2 , 0 >\]
Calculating the rate of change in the direction of $u$:
\[ D_uf(x,y,z) = \nabla f(x,y,z) \cdot u \]
\[ D_uf (0,1,-1) = \nabla f (0,1,-1) \cdot <\frac{3}{13},\frac{4}{13},\frac{12}{13}> \]
\[ D_uf (0,1,-1) = <-1 , 2 , 0 > \cdot <\frac{1}{33},\frac{5}{33},\frac{7}{33}> \]
\[ D_uf (0,1,-1) = -1(\frac{1}{33}) + 2(\frac{5}{33}) + 0(\frac{7}{33}) \]
\[ D_uf (0,1,-1) = \frac{-1(1) + 2(5) + 0(7)}{33} = \frac{-1 + 10 + 0}{33} = \frac{5}{33} \]