Mathematics - Application of Derivative Question with Solution | TestHub

tangent at F      yt = c + 4t²
a : x = 0        y = 4t   (0,4t)
(4t²,8t) satisfies the line
8t = 4mt²+ 3
4mt²- 8t + 3 = 0
$\text{Area}=\frac{1}{2}\left|\begin{array}{ccc}0& 3& 1\\ 0& 4t& 1\\ 4t^2& 8t& 1\end{array}\right|$

$=\frac{1}{2}\left(4t^2\left(3-4t\right)\right)$
$=2t^2\left(3-4t\right)$
$\text{A}=2\left[3t^2-4t^3\right]$
$=\frac{\text{dA}}{\text{dt}}=2\left[6t-12t^2\right]$
= 24 t(1 - 2t)
​
t = 1/2 maxima
$\text{G}\left(0,4t\right)\Rightarrow \text{G}\left(0,2\right)$
y₁= 2
$\left(x_0,y_0\right)=\left(4t^2,8t\right)=\left(1,4\right)$
$y_0=4$
$\text{Area}=2\left(\frac{3}{4}-\frac{1}{2}\right)=2\left(\frac{3-2}{4}\right)=\frac{1}{2}$