Question

Find an equation of the tangent line and normal line to the curve at the given point.

41. \(y = {x^4} + 2{e^x},\;\;\;\;\;\;\;\left( {0,2} \right)\)

Short answer

The required equation of tangent line is \(y = 2x + 2\) and equation of normal line is \(y = - \frac{1}{2}x + 2\).

Step-by-step solution

Find the slope of the equation

Determine the derivative \(\frac{{dy}}{{dx}}\) to represent the slope.

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{x^4} + 2{e^x}} \right)\\ &= \frac{d}{{dx}}\left( {{x^4}} \right) + \frac{d}{{dx}}\left( {2{e^x}} \right)\\y' &= 4{x^{4 - 1}} + 2{e^x}\\y' &= 4{x^3} + 2{e^x}\end{aligned}\)

Find the slope of the equation at given point

Substitute the given point into the obtained equation of slope to find the slope at particular point.

\(\begin{aligned}y'\left( 0 \right) &= 4{\left( 0 \right)^3} + 2{e^0}\\ &= 0 + 2\\ &= 2\end{aligned}\)

Determine the tangent line equation

The equation of tangent line is given by \(y - {y_1} = m\left( {x - {x_1}} \right)\), where \(\left( {{x_1},{y_1}} \right)\) is the given pint and \(m\) is the slope at given point.

Substitute the values into the equation to find the equation of tangent line at the point \(\left( {1,0} \right)\).

\(\begin{aligned}y - 2 &= 2\left( {x - 0} \right)\\y &= 2x + 2\end{aligned}\)

Thus, the required equation of tangent line is \(y = 2x + 2\).

Determine the Normal line equation

Since the slope of normal line is the negative reciprocal of slope of tangent lime. This implies that slope of normal line is \( - \frac{1}{2}\).

Substitute the slope and given point into the equation \(y - {y_1} = m\left( {x - {x_1}} \right)\) to find the equation of normal line.

\(\begin{aligned}y - 2 &= - \frac{1}{2}\left( {x - 0} \right)\\y &= - \frac{1}{2}x + 2\end{aligned}\)

Thus, the required equation of normal line is \(y = - \frac{1}{2}x + 2\).