Question

Find the gradient vector field for the scalar function. (That is, find the conservative vector field for the potential function.) $$ g(x, y, z)=x \arcsin y z $$

Short answer

The gradient vector field for the scalar function \(g(x, y, z) = x \arcsin yz\) is \(∇g= (\arcsin yz) \widehat{i} + (x . \frac{1}{\sqrt{1-y^2z^2}} . z) \widehat{j} + ( x . \frac{1}{\sqrt{1-y^2z^2}} . y ) \widehat{k}\).

Step-by-step solution

Find the partial derivative with respect to x

To begin, let's calculate the partial derivative of \(g(x, y, z)\) with respect to x. Since \(x\) is involved in a multiplication operation and the other factor \(\arcsin yz\) does not involve \(x\), we can use the simple derivative rule:\[\frac{\partial g}{\partial x} = \arcsin yz\]

Find the partial derivative with respect to y

Next, we calculate the partial derivative of \(g(x, y, z)\) with respect to \(y\) using the chain rule:\[\frac{\partial g}{\partial y} = x . \frac{1}{\sqrt{1-y^2z^2}} . z\]

Find the partial derivative with respect to z

Finally, we determine the partial derivative of \(g(x, y, z)\) with respect to \(z\) again using the chain rule:\[\frac{\partial g}{\partial z} = x . \frac{1}{\sqrt{1-y^2z^2}} . y\]

Write the Gradient Vector Field

The gradient vector field is obtained by packaging the partial derivatives into a vector. Therefore, the gradient vector field, often designated as \(∇g\) is:\[∇g= \frac{\partial g}{\partial x} \widehat{i} + \frac{\partial g}{\partial y} \widehat{j} + \frac{\partial g}{\partial z} \widehat{k} = (\arcsin yz) \widehat{i} + (x . \frac{1}{\sqrt{1-y^2z^2}} . z) \widehat{j} + ( x . \frac{1}{\sqrt{1-y^2z^2}} . y ) \widehat{k}\]