Question

Find an equation of the line that is both tangent to the curve \(y = {x^4} + 1\) and parallel to the line \(32x - y = 15\).

Short answer

The equation of the tangent line is \(y = 32x - 47\).

Step-by-step solution

 Tangent line to the curve

The tangent line to the curve \(y = f\left( x \right)\) at \(\left( {a,f\left( a \right)} \right)\) is the line through \(\left( {a,f\left( a \right)} \right)\) whose slope is equal to \(f'\left( a \right)\), the derivative of \(f\) at \(a\).

Find an equation of the line

The derivative of \(y = {x^4} + 1\) as shown below:

\(\begin{align}y' &= \frac{d}{{dx}}\left( {{x^4} + 1} \right)\\ &= \frac{d}{{dx}}\left( {{x^4}} \right) + \frac{d}{{dx}}\left( 1 \right)\\ &= 4{x^3}\end{align}\) ……(1)

Rewrite the line equation as shown below:

\(\begin{align}32x - y &= 15\\y &= 32x - 15\end{align}\)

Evaluate the slope of the line as shown below

\(\begin{align}y' &= \frac{d}{{dx}}\left( {32x - 15} \right)\\ &= 32\frac{d}{{dx}}\left( x \right) - \frac{d}{{dx}}\left( {15} \right)\\ &= 32 - 0\\ &= 32\end{align}\)

The slope of the line is 32. Therefore, any parallel line will also have a slope of 32.

Substitute \(y' = 32\) in the equation (1) as shown below:

\(\begin{align}4{x^3} &= 32\\{x^3} &= 8\\x &= 2\end{align}\)

It is the \(x - \) coordinates of the point on the curve where the slope is 32.

Substitute \(x = 2\) in the equation \(y = {x^4} + 1\) to determine \(y - \) coordinate as shown below:

\(\begin{align}y &= {2^4} + 1\\y &= 16 + 1\\y &= 17\end{align}\)

The equation of the tangent line at \(\left( {2,17} \right)\) as shown below:

\(\begin{align}y - 17 &= 32\left( {x - 2} \right)\\y &= 32x - 64 + 17\\y &= 32x - 47\end{align}\)

Thus, the equation of the tangent line is \(y = 32x - 47\).