Determine the equation of the tangent line to the curve $y = x\sqrt{x}$ which is parallel
to the line $y = 1 + 3x$
$
\begin{equation}
\begin{aligned}
\text{Given:}&& \text{Curve}\quad y &= x\sqrt{x}\\
\phantom{x}&& \text{Line} \quad y &= 1 + 3x
\end{aligned}
\end{equation}
$
The equation of the curve is parallel to the equation of the line which means they have the same slope.
From the formula $mx+b$, we take the equation of the line $y = 1+3x$
The slope$(m)$ is equals to 3.
$
\begin{equation}
\begin{aligned}
y &= x\sqrt{x}\\
\\
y'&= x \frac{d}{dx}(x^{\frac{1}{2}}) + (x)^{\frac{1}{2}} \frac{d}{dx}(x)
&& \text{Using Product Rule}\\
\\
y'&= (x) \left( \frac{1}{2\sqrt{x}} \right) + (x)^{\frac{1}{2}} (1)
&& \text{Simplify the equation}\\
\\
y'&= \frac{x}{2\sqrt{x}} + \sqrt{x}
&& \text{Get the LCD}\\
\\
y'&= \frac{x+2x}{2\sqrt{x}}
&& \text{Combine like terms}\\
\\
y'&= \frac{3x}{2\sqrt{x}}
\end{aligned}
\end{equation}
$
Let $y' =$ slope$(m)$
$
\begin{equation}
\begin{aligned}
m &= \frac{3x}{2\sqrt{x}}
&& \text{Substitute the value of slope}(m)\\
\\
3 &= \frac{3x}{2\sqrt{x}}
&& \text{Simplify the equation}\\
\\
3 &= \frac{3\sqrt{x}}{2}
&& \text{Multiplky both sides by } \frac{2}{3}\\
\\
\frac{6}{2} &= \sqrt{x}
&& \text{Simplify the equation}\\
\\
(\sqrt{x})^2 &= (3)^2
&& \text{Take the square of both sides}\\
\\
x &= 9
\end{aligned}
\end{equation}
$
Use the equation of the curve to get the value of $y$
$
\begin{equation}
\begin{aligned}
y &= x\sqrt{x} && \text{Substitute the value of } x\\
\\
y &= (9)(\sqrt{9}) && \text{Simplify the equation}\\
\\
y &= 27
\end{aligned}
\end{equation}
$
By using point slope form
$
\begin{equation}
\begin{aligned}
y - y_1 &= m (x-x_1)
&& \text{Substitute the value of } x, y \text{ and slope}(m)\\
\\
y - 27 &= 3(x-9)
&& \text{Distribute 3 in the equation.}\\
\\
y - 27 &= 3x - 27
&& \text{Add 27 to each sides}\\
\\
y &= 3x - 27 + 27
&& \text{Combine like terms}
\end{aligned}
\end{equation}
$
The equation of tangent line is $y = 3x$
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