Question

Find the equation of the tangent line to the curve $y=1+3x?$

Answer 1

In this question we have to find the equation of the tangent line to the curve which is parallels to another line. Use the curve equation and simplify it to take a differentiation in order to find the value of $x$ Use this value to find $'y'.$ then use the point slope formula to determine the equation of tangent.

Complete step by step solution:
The curve equation is $y=x\sqrt{x.}$ which is parallel to the line $y=1+3x.$ We know that parallel line means equal slope. The slope of the curve at $x=a$ is given by evaluating $f'(a),$ where $f(x)$ is the derivative.
We need to find the derivative, therefore first of all, the function can be rewritten as
$y=x{{\left( x \right)}^{{}^{1}/{}_{2}}}\,\,\,\,\,\,\,\,\,\,\,..........\left ( \sqrt{x}={{x}^{{}^{1}/{}_{2}}} \right)$
We can now use the power rule. We get
$y={{x}^{{}^{3}/{}_{2}}}$
Now, differentiate above equation with respect to $'x'$ we get,
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{{}^{3}/{}_{2}}} \right)$
Then by using formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ where $n$ is any real number above expression can be written as,
$\dfrac{dy}{dx}=\dfrac{3}{2}{{x}^{{}^{3}/{}_{2}-1}}$
Now, after subtracting $1$ from $\dfrac{3}{2}$ above expression will written as,
$\dfrac{dy}{dx}=\dfrac{3}{2}{{x}^{{}^{1}/{}_{2}}}$
By comparing the given line $y=1+3x$ with $y=mx+c$ where $m$ is slope and $c$ is intercept, the slope of the line is $3.$ so we set $\dfrac{dy}{dx}$ to 3 and solve for $x.$
$3=\dfrac{3}{2}{{x}^{{}^{1}/{}_{2}}}$
By dividing with $\dfrac{3}{2}$ to both side of above equation we have,
$\dfrac{\dfrac{3}{3}}{2}={{x}^{{}^{1}/{}_{2}}}$
Now cancel the common factor
$2={{x}^{{}^{1}/{}_{2}}}$
Here, we know that ${{4}^{{}^{1}/{}_{2}}}=2$ therefore the value of $x$ is,
$x=4$
We now use this point to determine the corresponding $y-$co$-$ordinate $.$
$y=x\sqrt{x}$
Put the derive value $x=4$ in above expression we have,
$y=4\sqrt{4}$
Since, the value of $\sqrt{4}$ is 2, above expression can be written as
$y=4\left( 2 \right)$
$y=8$
From this, we know the slope and point of contact. The equation is given by $:$
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
$y-8=3\left( x-4 \right)$
$y-8=3x-12$
$y=3x-4$
Hence, the equation of the tangent line to the curve $y=x\sqrt{x}$ that is parallel to the line $y=1+3x$ is $y=3x+4.$

Note: Find the first derivative of $f(x).$ Then plug $x$ value of the indicated point into $f'(x)$ to find the slope at $x.$ Also, plug $x$ value into $f(x)$ to find the $y-$coordinate of the tangent point. Combine the slow and point using the point slope formula to find the equation for the tangent line.