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the paraboloid z=4-2(x²)-(y²) intersects the plane y=1 in a parabola. find an equation the line tangent to this parabola at the point (3,1,-15).

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Final answer:

To find the equation of the tangent line to the parabola at the point (3, 1, -15), we differentiate the equation with respect to x to find the slope. Substituting the given point into the derivative, we find the slope of the tangent line. Using the point-slope form of a line, we can write the equation of the tangent line as y = -12x + 37.

Step-by-step explanation:

To find the equation of the line tangent to the parabola at the point (3, 1, -15), we first need to find the derivative of the parabola. The given parabola is z = 4 - 2x² - y². To find the slope at the point (3, 1, -15), we differentiate the equation with respect to x and then substitute x = 3 and y = 1. The derivative of z with respect to x is dz/dx = -4x. Substituting x = 3, we get dz/dx = -12.

Next, we find the slope of the tangent line by using the derivative dz/dx = -12. Since the line is tangent to the parabola, the tangent line will have the same slope as the derivative. Therefore, the slope of the tangent line is -12.

Using the point-slope form of a line, which is y - y₁ = m(x - x₁) where (x₁, y₁) is the given point and m is the slope, we can write the equation of the tangent line as y - 1 = -12(x - 3). Simplifying the equation, we have y = -12x + 37.

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