Question

At what point on the curve \(y = \sqrt {1 + 2x} \) is the tangent line perpendicular to the line \(6x + 2y = 1\)?

Short answer

The points on the curve on which tangent line is perpendicular to given line is \(\left( {4,3} \right)\).

Step-by-step solution

Formulas

Chain Rule: \(\frac{d}{{dx}}\left( {{u^n}} \right) = n{u^{n - 1}}\frac{{du}}{{dx}}\)

When two lines are perpendicular to each other, then the relation between their slopes is given by\({m_1} \cdot {m_2} = - 1\), where \({m_1}\) and \({m_1}\) are slope of lines.

Differentiate the given curve

Differentiate \(y = \sqrt {1 + 2x} \)with respect to \(x\) by using chain rule.

\(\begin{aligned}y' &= \frac{d}{{dx}}\left( {\sqrt {1 + 2x} } \right)\\ &= \frac{d}{{dx}}{\left( {1 + 2x} \right)^{\frac{1}{2}}}\\ &= \frac{1}{2}{\left( {1 + 2x} \right)^{\frac{1}{2} - 1}} \cdot \frac{d}{{dx}}\left( {1 + 2x} \right)\\ &= \frac{1}{2}{\left( {1 + 2x} \right)^{ - \frac{1}{2}}} \cdot 2\\ &= \frac{1}{{\sqrt {1 + 2x} }}\end{aligned}\)

Find slope of the given line \(6x + 2y = 1\)

Given equation of line is\(6x + 2y = 1\).

Isolate \(y\) from the equation.

\(y = - 3x + \frac{1}{2}\)

Which is the point slope form of a line. Where slope is \( - 3\), that is, \({m_1} = - 3\).

Slope of the tangent line perpendicular to the given line

Let slope of the perpendicular line, which is perpendicular to \(6x + 2y = 1\) be \({m_2}\).

According to the relation between two perpendicular lines, \({m_1} \cdot {m_2} = - 1\).

Then,

\(\begin{aligned} - 3 \cdot {m_2} &= - 1\\{m_2} &= \frac{{ - 1}}{{ - 3}}\\ &= \frac{1}{3}\end{aligned}\)

And the obtained slope is the slope of tangent line, so \(y' = \frac{1}{3}\).

\(\frac{1}{3} = \frac{1}{{\sqrt {1 + 2x} }}\)

Find the value of \(x\) from the obtained equation.

\(\begin{aligned}\frac{1}{3} &= \frac{1}{{\sqrt {1 + 2x} }}\\\sqrt {1 + 2x} &= 3\\1 + 2x &= 9\\2x &= 8\\x &= 4\end{aligned}\)

Find point on the given curve

Substitute \(x = 4\) into the given curve \(y = \sqrt {1 + 2x} \) and solve for \(y\).

\(\begin{aligned}y &= \sqrt {1 + 2\left( 4 \right)} \\ &= \sqrt {1 + 8} \\ &= \sqrt 9 \\ &= 3\end{aligned}\)

Hence, the point on the curve on which tangent line is perpendicular to given line is \(\left( {4,3} \right)\).