91Ó°ÊÓ

To find the gradient of \(f(x,y)\), we will compute its partial derivatives with respect to \(x\) and \(y\). For \(f_x\): \(\frac{\partial f}{\partial x}=-\frac{1}{4}x\) For \(f_y\): \(\frac{\partial f}{\partial y}=-\frac{1}{16}y\) Now, we have the gradient vector \(\nabla f(x,y) = \left\langle-\frac{1}{4}x, -\frac{1}{16}y\right\rangle\)

Now, we will find the gradient at point \(P(4,0)\). \(\nabla f(4,0)=\left\langle -\frac{1}{4}(4), -\frac{1}{16}(0) \right\rangle=\left\langle -1, 0 \right\rangle\)

The level curve is given by the equation \(f(x, y) = 12\), where the function \(f(x, y) = 16 - \frac{x^2}{4} - \frac{y^2}{16}\). By plugging in the function, we get: \(12 = 16 - \frac{x^2}{4} - \frac{y^2}{16}\) We need to find the gradient of this level curve, which requires computing its partial derivatives with respect to \(x\) and \(y\). For \(f_x\): \(\frac{\partial}{\partial x} (16-\frac{x^2}{4}-\frac{y^2}{16}) = -\frac{1}{2}x\) For \(f_y\): \(\frac{\partial}{\partial y} (16-\frac{x^2}{4}-\frac{y^2}{16}) = -\frac{1}{8}y\) Now, we have the gradient vector of the level curve \(\nabla g(x,y) = \left\langle -\frac{1}{2}x, -\frac{1}{8}y\right\rangle\)

Now, we will find the gradient at point \(P(4,0)\). \(\nabla g(4,0)=\left\langle -\frac{1}{2}(4), -\frac{1}{8}(0) \right\rangle=\left\langle -2, 0 \right\rangle\)

To find the slope of the tangent line, we need to remember that the tangent line is orthogonal to the gradient vector when their dot product is equal to zero. \(\nabla f(x,y) \cdot \nabla g(x,y)=(-1)(-2)+(0)(0) = 2\) As their dot product is not zero, these gradients are not orthogonal. Hence, there must be an error in the question or given information since it is stated that tangent line should be orthogonal to the gradient. Note: The slope of the tangent line can also be expressed as the negative reciprocal of the slope of the function's gradient. For example, if a gradient vector had y = mx, the slope of the tangent would be -1/m. Since the gradients are not orthogonal in this problem, we cannot find the slope.