The area (in sq. units) of the region bounded by the curve $\sqrt x+\sqrt y=1,x,y\ge0$, and the tangent to it at the point $(\frac14,\frac14)$ is : $\frac1{24}/\frac1{8}/\frac1{36}/\frac1{12}$?
To start this, I need to make graph of $\sqrt x+\sqrt y=1$, I guess. But not able to do so. It's neither circle nor ellipse. How to approach?
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$\begingroup$Above is the plot of the curve.
so the area from $x \in [0,1]$ is $$A(x) = \int_{0}^{1}(1-\sqrt{x})^2 dx$$You can easily see that the required area R(x) is $$R(x) = I(x) - \text{Area of triangle}$$
You can see from the graph that the area of triangle bounded by the tangent and the $x$ axis and the $y$ axis is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$. The integral $I(x) = \frac{1}{6}$Hence, $R(x) = \frac{1}{24}$
Edited the answer I had some misconceptions about bounding the curve
So it is not possible to exactly plot the curve but it is possible to plot the tangent and plot a rough curve to get the above idea. The key thing is that the curve passes through $(0,1),(1/4,1/4),(1,0)$. And the tangent intersects the $y$ and the $x$ -axis at $(0,1/2)$ and $(1/2,0)$. From this info I draw the following.
I think we can do the rest from this image too. You get the limits of the integration, you get the area of the triangle to be subtracted. To do this under one minute will still be hard for me at least.
$\endgroup$ 2 $\begingroup$$$ \int_{0}^{1}(1-\sqrt{x})^2\,dx - \int_{0}^{1/2}\left(\frac{1}{2}-x\right)\,dx $$is the wanted area (can you figure out why?), so the answer is given by$$ 1-\frac{4}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}=\color{red}{\frac{1}{24}}. $$
$\endgroup$ 3 $\begingroup$The graph is an arc of parabola with line bissector (y=x) as its axis of symmetry. Here is why.
Setting $\sqrt{x}=t \iff x=t^2$, we have $y=1-\sqrt{x}=1-t \iff y^2=t^2-2t+1$
Using rotation by $+\pi/4$ combined with homothety with ratio $\sqrt{2}$ :
$$\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}1&-1\\1& \ \ 1\end{pmatrix}\begin{pmatrix}t^2\\t^2-2t+1\end{pmatrix}=\begin{pmatrix}2t-1\\2t^2-2t+1\end{pmatrix}$$
from which, plugging $t=\frac{(X+1)}{2}$ in the expression of $Y$, you get $Y$ as a quadratic in $X$.
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