Use implicit differentiation to find the equation of the tangent line to the curve 𝑥2 + 𝑥𝑦 + 𝑦2 = 3 at the point (1,1).
Question
Use implicit differentiation to find the equation of the tangent line to the curve 𝑥2 + 𝑥𝑦 + 𝑦2 = 3 at the point (1,1).
Solution
首先,我们需要对方程 x^2 + xy + y^2 = 3 进行隐式微分。这意味着我们需要对方程的每一部分分别求导。
对于 x^2,它的导数是 2x。 对于 xy,我们需要使用乘法法则,它的导数是 x * dy/dx + y。 对于 y^2,它的导数是 2y * dy/dx。
所以,我们得到的微分方程是 2x + x * dy/dx + y + 2y * dy/dx = 0。
然后,我们需要将 dy/dx 的项集中到一边,其他的项移到另一边。这样我们可以得到 dy/dx = -(2x + y) / (x + 2y)。
现在,我们需要找到点 (1,1) 处的斜率。我们将 x 和 y 都替换为 1,得到 dy/dx = -(21 + 1) / (1 + 21) = -1。
所以,我们知道在点 (1,1) 处,曲线的斜率是 -1。
最后,我们需要找到切线的方程。切线的斜率是 -1,而且我们知道切线在点 (1,1) 处与曲线相交,所以切线的方程是 y - 1 = -1 * (x - 1)。
简化这个方程,我们得到 y = -x + 2。这就是在点 (1,1) 处,曲线 x^2 + xy + y^2 = 3 的切线的方程。
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