Question

Find the equation of the tangent line to the curve \[y=1+3x?\]

Answer 1

Hint: 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.