Question

The parabola shown below has vertex at the origin, and passes through the point (3,2). The shaded region is bounded by the parabola, its tangent line at the point (3,2), and the x axis. Find its area.

Thanks for the help!

Answer 1

Answer:

0.5 square units

Step-by-step explanation:

Find the equation of the parabola in the form $y=ax^2.$ This parabola passes through the point (3,2), so

$2=a\cdot 3^2\\ \\2=9a\\ \\a=\dfrac{2}{9}$

Thus, the parabola equation is

$y=\dfrac{2}{9}x^2$

Now, find the equation of the tangent line.

$y'=\left(\dfrac{2}{9}x^2\right)'=\dfrac{2}{9}\cdot 2x=\dfrac{4}{9}x\\ \\y'(3)=\dfrac{4}{9}\cdot 3=\dfrac{4}{3}$

The equation of the tangent line is

$y-y(x_0)=y'(x_0)(x-x_0)\\ \\y-2=\dfrac{4}{3}(x-3)\\ \\y=\dfrac{4}{3}x-2$

This line intersects x-axis at point (1.5, 0), because

$\dfrac{4}{3}x-2=0\\ \\x=2\cdot \dfrac{3}{4}=1.5$

Find the shaded area:

$A=\int\limits_0^{1.5}\dfrac{2}{9}x^2dx+\int\limits_{1.5}^3\left(\dfrac{2}{9}x^2-\dfrac{4}{3}x+2\right)dx=\left(\dfrac{2}{9}\cdot \dfrac{x^3}{3}\right)|_0^{1.5}+\left(\dfrac{2}{9}\cdot \dfrac{x^3}{3}-\dfrac{4}{3}\cdot \dfrac{x^2}{2}+2x\right)|_{1.5}^3=\\ \\=\dfrac{2}{27}\cdot (1.5^3-0^3)+\dfrac{2}{27}\cdot (3^3-(1.5)^3)-\dfrac{2}{3}\cdot(3^2-1.5^2)+2\cdot (3-1.5)=$

$=\dfrac{2}{27}\cdot \dfrac{27}{8}+\dfrac{2}{27}\cdot \left(27-\dfrac{27}{8}\right)-\dfrac{2}{3}\cdot \left(9-\dfrac{9}{4}\right)+2\cdot \dfrac{3}{2}=\\ \\=\dfrac{1}{4}+2-\dfrac{1}{4}-6+\dfrac{3}{2}+3=\\ \\=0.5\ un^2.$