Tangents and gradients

Calculus Level 3

For $-1 \lt x \lt 1, -1 \lt y \lt 1$ satisfying $x^2 + y^2 = 1$ , what is the range of all the possible values of $\dfrac{dy}{dx}$ ?

\mathbb R^+

\mathbb R

\mathbb R^-

\mathbb R \backslash \{ 0 \}

1 solution

Alex G
Aug 14, 2016

Remember that $\dfrac{dy}{dx}$ is the slope of the tangent line at a point $(x,y)$ .

Now, the graph of $x^2+y^2 =1$ is a circle centered at the origin with radius 1. The slope of the line tangent to the circle varies continuously through the real numbers, going from positive infinity to negative infinity.

However, the domain of the derivative (i.e. the slope of the tangent line) does not include $x \ , \ y = \pm 1$ . At the four points on the circle where this occurs, $(1,0),(-1,0),(0,1),(0,-1)$ , the derivative (the slope of the tangent line) is either 0 or undefined.

Therefore, the range of the derivative is all real numbers except 0.

Tangents and gradients

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