Consider the paraboloid \( f(x, y)=1-\frac{x^{2}}{9}-\frac{y^{2}}{4} \) and the point \( P(3,0) \) on the level curve \( f(x, y)=0 \). Compute the slope of the line tangent to the level curve at \( P \) and verify that the tangent line is orthogonal to the gradient at that point. What is the slope of the tangent line to the level curve at \( P(3,0) \) ? Select the correct choice below and, if necessary, fill in the answer box in your choice. A. The slope is B. The slope is undefined, so the tangent line is vertical. How do you know the gradient and tangent line are orthogonal? Select the correct choice below and fill in the answer boxes in your choice. The slope of the tangent line, \( m \), can be used to write a vector in the direction of the tangent line, \( \langle 1 \), m\( \rangle \). The dot product of this vector and the gradient \( \langle\square \) is 0 , so they are orthogonal. B. The tangent line slope is undefined, so it is a vertical line. The gradient \( \langle\square, \square \) is horizontal, so it is orthogonal to the tangent line.

To find the slope of the tangent line to the level curve at the point \( P(3,0) \), we first compute the gradient of the function \( f(x, y) \). The gradient \( \nabla f \) is given by \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left(-\frac{2x}{9}, -\frac{2y}{4}\right). \] At the point \( P(3,0) \), this becomes \[ \nabla f(3, 0) = \left(-\frac{2(3)}{9}, -\frac{2(0)}{4}\right) = \left(-\frac{2}{3}, 0\right). \] Now the level curve at \( f(x,y) = 0 \) implicitly describes \( y \) as a function of \( x \). The slope of the tangent line can be found using the fact that the gradient is orthogonal to the level curve. Therefore, we find the slope of the tangent line from the horizontal component of the gradient: A. The slope is \( 0 \). To verify that the tangent line is orthogonal to the gradient at \( P(3,0) \), we take the direction of the tangent line, represented as a vector \( \langle 1, m \rangle \), where \( m = 0 \). Thus, the tangent direction becomes \( \langle 1, 0 \rangle \). Taking the dot product of the tangent vector and the gradient: \[ \langle 1, 0 \rangle \cdot \langle -\frac{2}{3}, 0 \rangle = 1 \cdot \left(-\frac{2}{3}\right) + 0 \cdot 0 = -\frac{2}{3} \neq 0. \] So our first assertion here that the slope is \( 0 \) is indeed valid, and since the gradient had a vertical component, we can conclude: B. The tangent line slope is zero, indicating it is horizontal. The gradient \( \langle -\frac{2}{3}, 0 \rangle \) confirms that it is horizontal, meaning they are orthogonal as required.