Techniques of Differentiation

Techniques of Differentiation - Exercise 4

Exercise 4. Find the points on the graph of y = x^3/2 - x^1/2 at which the tangent line is parallel to the line y+2x = 1. Also find the points on the same graph at which the tangent line is perpendicular to the line y-x = 3.

Answer. First let us find the points on the graph at which the tangent line is parallel to the line y+2x = 1. For that, we need the slope of any tangent line which is given by the derivative

$\begin{displaymath}\frac{dy}{dx} = \frac{3}{2} x^{1/2} - \frac{1}{2} x^{-1/2} = \frac{3x-1}{2 \sqrt{x}} \cdot\end{displaymath}$

We know that two lines are parallel if and only if they have the same slope. Since the slope of the line y+2x = 1 is -2, we then are left to solve

$\begin{displaymath}\frac{dy}{dx} = \frac{3x-1}{2 \sqrt{x}} = -2 \;.\end{displaymath}$

We rewrite this equation to get

$\begin{displaymath}3x + 4 \sqrt{x} -1 = 0\end{displaymath}$

which is a quadratic equation in $\sqrt{x}$ . Therefore we must have

$\begin{displaymath}\sqrt{x} = \frac{-2 \pm \sqrt{7}}{3} \;\cdot\end{displaymath}$

Since $\sqrt{x}$ has to be non-negative, we have to discard the negative solution and are thus left with

$\begin{displaymath}\sqrt{x} = \frac{-2 + \sqrt{7}}{3} \end{displaymath}$

or equivalently

$\begin{displaymath}x = \left( \frac{-2 + \sqrt{7}}{3} \right)^2 \;.\end{displaymath}$

So there is only one point on the graph at which the tangent line is parallel to the line y+2x = 1.

Next we look for the points on the graph at which the tangent line is perpendicular to the line y-x = 3. In this case, the slope of tangent line should be

$\begin{displaymath}- \frac{1}{1/2} = -1 \;.\end{displaymath}$

As before we must solve the equation

$\begin{displaymath}\frac{dy}{dx} = \frac{3x-1}{2 \sqrt{x}} = -1 \;.\end{displaymath}$

We rewrite this equation to get

$\begin{displaymath}3x + 2 \sqrt{x} -1 = 0\;.\end{displaymath}$

Therefore

$\begin{displaymath}\sqrt{x} = \frac{-1 \pm \sqrt{4}}{3} = \frac{-1 \pm 2}{3}\;\cdot\end{displaymath}$

which implies $\sqrt{x} = \displaystyle \frac{1}{3}$ or equivalently

$\begin{displaymath}x = \frac{1}{9}\;.\end{displaymath}$

This is the only point on the graph at which the tangent line is perpendicular to the line y-x = 3.

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