At the given point, find the line that is normal to the curve at the given point.

6x2y - π cos y = 7π, normal at (1, π)

Question

At the given point, find the line that is normal to the curve at the given point.

6x2y - π cos y = 7π, normal at (1, π)

BlackTom AI · Accepted Answer

To begin, I’ll differentiate the given implicit curve to find the slope of the tangent at a general point. The curve is 6x^2 y − π cos y = 7π. Differentiating both sides with respect to x yields: 12xy + 6x^2 y' − (−π sin y) y' = 0, which simplifies to 12xy + (6x^2 + π sin y) y' = 0. Solving for y' gives y' = −12xy / (6x^2 + π sin y).

Next, we substitute the point (x, y) = (1, π) into the derivative. Since sin π = 0, the expression becomes y' = −12·1·π / (6·1^2 + π·0) = −12π / 6 = −2π. This is the slope of the tangent line at the given point.

A normal line is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent slope. The tangent slope is −2π, hence the normal slope is +1/(2π).

Now form the equation of the line with slope 1/(2π) passing through (1, π). Using y − y0 = m(x − x0): y − π = (1/(2π))(x − 1). Distributing gives y = (1/(2π))x − 1/(2π) + π. This is the explicit equation of the normal line in slope-intercept form: y = (1/(2π))x + π − 1/(2π).

Turning to the provided options, observe that each option presents a line with a different slope or a malformed constant term. For clarity:
- The first option shows a slope of −1, which does not match the computed normal slope of +1/(2π).
- The second option shows a slope of +1, which again does not match +1/(2π).
- The third option shows a slope of −2π, which corresponds to the tangent line, not the normal.
- The fourth option also presents a line with a slope that does not equal +1/(2π).

Thus, none of the given choices matches the correct normal line equation y = (1/(2π))x + π − 1/(2π).

At the given point, find the line that is normal to the curve at the given point. 6x2y - π cos y = 7π, normal at (1, π) 单项选择题

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