Question

(a). Find the slope of the tangent to the curve \(y = 2\sqrt x \)at the point where\(x = a\).

(b). Find equations of the tangent lines at the points \(\left( {1,2} \right)\) and \(\left( {9,6} \right)\).

(c). Graph the curve and both tangents on a same screen.

Short answer

a. The slope of the tangent line is \(m = \frac{1}{{\sqrt a }},\,\,\forall \,\,a > 0\).

b.The equations for tangent lines are:

\(\begin{array}{l}y = x + 1\\y = \frac{1}{3}x + 3\end{array}\).

c. The graph for curve and tangent lines is:

Step-by-step solution

Slope of the Function

The slope of any function can be evaluated using the method of differentiation for all finite points on graph at which tangent to that curve can be projected in that plane itself.

Step 2:Find the slope of the given function (a)

The function is\(y = 2\sqrt x \).

Now, for point where \(x = a\), we have:

\(\begin{aligned}m &= \mathop {\lim }\limits_{x \to a} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ &= \mathop {\lim }\limits_{x \to a} \frac{{\left( {2\sqrt x - 2\sqrt a } \right)}}{{x - a}}\\ &= \mathop {\lim }\limits_{x \to a} \frac{{\left( {2\sqrt x - 2\sqrt a } \right)\left( {2\sqrt x + 2\sqrt a } \right)}}{{\left( {x - a} \right)\left( {2\sqrt x + 2\sqrt a } \right)}}\\ &= \mathop {\lim }\limits_{x \to a} \frac{{4x - 4a}}{{\left( {x - a} \right)\left( {2\sqrt x + 2\sqrt a } \right)}}\\ &= \mathop {\lim }\limits_{x \to a} \frac{4}{{\left( {2\sqrt x + 2\sqrt a } \right)}}\\ &= \frac{4}{{\left( {2\sqrt a + 2\sqrt a } \right)}}\\ &= \frac{1}{{\sqrt a }}\end{aligned}\)

Hence, the slope of the tangent line is \(m = \frac{1}{{\sqrt a }},\,\,\forall \,\,a > 0\).

Find the equations of the tangent linesof the function at given points (b)

Since the function is \(y = 2\sqrt x \)and the slope is \(m = \frac{1}{{\sqrt a }}\), the tangent line equation at point (1, 2) is given by:

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

Also, thetangent line equation at point (9, 6) is given by:

\(\begin{aligned}\left( {y - 6} \right) &= \left( {\frac{1}{{\sqrt a }}} \right)\left( {x - 9} \right)\\y - 6 &= \left( {\frac{1}{{\sqrt 9 }}} \right)\left( {x - 9} \right)\\y &= \frac{1}{3}x + 3\end{aligned}\)

Hence, theequations for tangent lines are:

\(\begin{array}{l}y = x + 1\\y = \frac{1}{3}x + 3\end{array}\).

.

Sketch the graph for curve and tangent lines (c)

Now, for the given function\(y = 2\sqrt x \), the graph for the curve and the tangent lines can be drawn as:

Hence, this is the required graph.