Question

Consider the vector field f(x,y) = (y/2, -x/2) and the curve c: y - x² = 1.

Determine the points along c at which f is tangent to c.

Answer 1

To determine the points along the curve where the vector field is tangent, we differentiate the equation of the curve to find its slope, then compare the tangent vector to the vector field.

Step-by-step explanation:

To determine the points along the curve where the vector field is tangent, we need to find the points where the tangent vector of the curve is parallel to the vector field.

First, differentiate the equation of the curve to find its slope. Using implicit differentiation, we have dy/dx = 2x.

The tangent vector to the curve is then given by T = (1, dy/dx) = (1, 2x).

Next, we compare the tangent vector to the vector field. Since the vector field is given by F = (y/2, -x/2), we can see that the x-components are equal: y/2 = 1. Solving for y, we get y = 2. Hence, the points along the curve where the vector field is tangent are (2, 2).