Answer to Question #114628 in Calculus for ANJU JAYACHANDRAN

Question

Find the slope of the curve y^2-x+1=0 at the points (2, -1) and (2, 1).

Answer 1

The graph of the function $y^2-x+1=0$ is a parabola, therefore, we rewrite the equation in the form $y= \sqrt{x-1}$ and $y=-\sqrt{x-1}$ . Since these graphs are symmetric with respect to Ox, we will consider the first option. The angle of inclination of the curve = the angle of inclination of the tangent, therefore, we find the tangent by the formula $y=f^\prime(x_0)(x-x_0)+f(x_0); x_0=2;f(x_0)=1; f^\prime(x)=\frac{1}{2\sqrt{x-1}}\to\\f^\prime(x_0)=\frac{1}{2}$

Then the tangent equation takes the form $y=\frac{1}{2}(x-2)+1$ or $y=\frac{1}{2}x$, where $\frac{1}{2}$ is the coefficient of the angle of inclination of the tangent, since the graphs are symmetrical, it means the tangents are also symmetrical. From this it follows that the slope coefficient of the tangent to the graph $y= \sqrt{x-1}$ is $\frac{1}{2}$ , and the slope coefficient to the tangent to the graph $y= -\sqrt{x-1}$ is $-\frac{1}{2}.$.

If you need to translate the coefficients into radians, then knowing that the slope and the angle are related with a help of the formula $k=\tan{\phi}$ , therefore, having solved the equations $\frac{1}{2}=\tan{\phi}$ , we find that the angle is $0.46$ radians. For another tangent $-0.46$ radians.