What is the slope of the tangent line of #x^2 + y^2 = 1/2 # at #x=1#?

Question

Charles Arocha · Accepted Answer

There is no point on the graph with #x=1#, so there is no tangent line at #x=1#.

For #x^2+y^2=1/2#, if #x=1#, then #y^2 = -1/2#. There is no real #y# that solves this. Geometrically The graph of #x^2+y^2 = 1/2# is a circle of radius #sqrt2/2#. The maximum #x# value on that circle is #x=sqrt2/2 ~~ 0.7071#.

Ezra Adams · Answer

undefined To find the slope of the tangent line at a specific point on a curve, you first need to find the derivative of the curve with respect to x, and then evaluate it at the given x-coordinate. The derivative of the curve $x^2 + y^2 = \frac{1}{2}$ with respect to x is $y' = -\frac{x}{y}$. Substituting $x = 1$ into this derivative, we get $y' = -\frac{1}{y}$. To find the value of y at $x = 1$, we can use the equation of the curve: $x^2 + y^2 = \frac{1}{2}$ and solve for y. Solving this equation yields two possible values for y: $y = \sqrt{\frac{1}{2} - 1}$ and $y = -\sqrt{\frac{1}{2} - 1}$. Substituting these values into $y'$, we get two slopes: $m_1 = -\sqrt{2} $ and $m_2 = \sqrt{2} $. Therefore, the slopes of the tangent lines at $x = 1$ are $ -\sqrt{2} $ and $ \sqrt{2} $.