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# Tutor profile: Ishika B.

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Ishika B.
University student and a mathematician in making.
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## Questions

### Subject:Calculus

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Question:

What are the dimensions of the rectangle with the largest area that can be inscribed in a semicircle of radius $3$?

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Ishika B.

with radius $3$ and center at $(0,0)$ given by\\ $x^2+y^2=32$\\ Solve the above equation for $y$\\ $y^2=9-x^2$\\ $y=\pm\sqrt{9-x^2}$\\ The equation for the upper semi cricle is $y=\sqrt{9-x^2}$\\ A point on the semi circle with $x$ coordinate $x$ has a $y$ coordinate equal to $y=\sqrt{9-x^2}$\\ The rectangle has a length $L=2x$ and a width (or height) $W=\sqrt{9-x^2}$\\ The area $A$ of the rectangle is given by\\ $A(x)=2x(\sqrt{9-x^2})$ , $0\leq x \leq 3$\\

### Subject:Partial Differential Equations

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Question:

Solve the boundary value problem\\ $\frac{dy}{dx}+16y=0\\ y(0)=-4\\ y(\pi /2)=5$

Inactive
Ishika B.

The general solution for this type of derivative is \\ $\frac{dy}{dx}+ay=0$\\ $y(x)=c_1\cos(\sqrt x)+c_2\sin(\sqrt a x)$\\ Therefore, the equation becomes\\ $y(x)=c_1 \cos(4x)+c_2 \sin(4x)$\\ Now we apply the boundary condition to solve for $c_1$ and $c_2$\\ $-4=y(0)=c_1$\\ $5=y(\pi /2)=c_2$\\ The solution will be \\ $y(x)=-4\cos(4x)+5\sin(4x)$

### Subject:Differential Equations

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Question:

Find the derivative of $x^3+5(x^2)$ at $x=2$.

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Ishika B.

$\frac{d(x^3+5(x^2))}{dx} = 3(x^2)+10x$\\ At $x$=$2$ ( We will put the value of $x$ as $2$ )\\ $3*(2^2)+10*2$=$32$

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