Question

# I) Find the directional derivative of π$π₯, π¦$ = π₯ sin π¦ at $1,0$ in the direction of the unit vector that make an angle of π/4 with positive π₯-axis.

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## Answer to a math question I) Find the directional derivative of π$π₯, π¦$ = π₯ sin π¦ at $1,0$ in the direction of the unit vector that make an angle of π/4 with positive π₯-axis.

Bud
4.6
To find the directional derivative of a function, we need to calculate the dot product of the gradient vector of the function and the unit vector in the given direction.

The gradient vector of the function f$x, y$ is given by the partial derivatives of f with respect to each variable.

Let's find the partial derivatives of f$x, y$:

\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} $x \sin y$ = \sin y

\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} $x \sin y$ = x \cos y

Now, we can find the gradient vector:

\nabla f = \left$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right$ = \left$\sin y, x \cos y\right$

The unit vector that makes an angle of π/4 with the positive π₯-axis is:

\mathbf{u}=\left$\cos\frac{\pi}{4},\sin\frac{\pi}{4}\right$=\left$\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right$

Now, we can calculate the dot product of the gradient vector and the unit vector:

\nabla f\cdot\mathbf{u}=\left$\sin y,x\cos y\right$\cdot\left$\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right$=\frac{\sqrt{2}}{2}\sin y+\frac{\sqrt{2}}{2}x\cos y

Substituting the coordinates of the point $1, 0$ into the equation above:

\nabla f\cdot\mathbf{u}=\frac{\sqrt{2}}{2}\cdot\sin0+\frac{\sqrt{2}}{2}\cdot1\cdot\cos0=\frac{\sqrt{2}}{2}\cdot0+\frac{\sqrt{2}}{2}\cdot1\cdot1=\frac{\sqrt{2}}{2}

Answer: The directional derivative of f$x, y$ = x sin y at $1, 0$ in the direction of the unit vector that makes an angle of π/4 with the positive π₯-axis is \frac{\sqrt{2}}{2} .

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