Geometrical Applications of Differential Equations

Q. If length of tangent at any point on the curve $y=f\left(x\right)$ intercepted between the point and the $x-$ axis is of length $1$ units, then the equation of the curve is:

1. $\ln \left|\frac{1-\sqrt{1-y^2}}{y}\right|+\sqrt{1-x^2}=\pm x+c$
2. $\ln \left|\frac{1-\sqrt{1-x^2}}{x}\right|+\sqrt{1-y^2}=\pm y+c$
3. $\ln \left|\frac{1-\sqrt{1-y^2}}{y}\right|+\sqrt{1-y^2}=\pm x+c$
4. $\ln \left|\frac{1-\sqrt{1-x^2}}{x}\right|+\sqrt{1-x^2}=\pm y+c$

Q. If the independent variable $x$ is changed to $y$, then the differential equation $x\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^3-\left(\frac{dy}{dx}\right)=0$ is changed to $x\frac{d^{2}x}{d{y}^2}+{\left(\frac{dx}{dy}\right)}^2=k$ where $k$ equals

Q. Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth $y$, where the constant of proportionality $k>0$ depends on the acceleration due to gravity and the geometry of the hole. If $t$ is measured in minutes and $k=\frac{1}{15}$, then the time to drain the tank if the water is $4\text{m}$ deep to start with is

1. $30\text{min}$
2. $45\text{min}$
3. $60\text{min}$
4. $80\text{min}$

Q. The equation of a curve passing through $(1,0)$ for which the product of the abscissa of a point $P$ and the intercept made by a normal at $P$ on the $x$-axis equals twice the square of the distance of the point $P$ from the origin $O$, is

1. $x^2+y^2=x^4$
2. $x^2+y^2=2x^4$
3. $x^2+y^2=4x^4$
4. $x^2+y^2=8x^4$

Q. Let $f\left(x\right)$ be a non-positive continuous function and $F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt\forall x\ge 0$ and $f\left(x\right)\ge cF\left(x\right)$ where $c>0$ and let $g:[0,\infty )\to R$ be a function such that $\frac{dg\left(x\right)}{dx}<g\left(x\right)\forall x>0$ and $g\left(0\right)=0.$
The solution set of inequality $g\left(x\right)({\cos }^{-1}x-{\sin }^{-1}x)\le 0$

1. $[-1,\frac{1}{\sqrt{2}}]$
2. $[\frac{1}{\sqrt{2}},1]$
3. $[0,\frac{1}{\sqrt{2}}]$
4. $(0,\frac{1}{\sqrt{2}}]$

Q. Let $f\left(x\right)$ be a non-positive continuous function and $F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt\forall x\ge 0$ and $f\left(x\right)\ge cF\left(x\right)$ where $c>0$ and let $g:[0,\infty )\to R$ be a function such that $\frac{dg\left(x\right)}{dx}<g\left(x\right)\forall x>0$ and $g\left(0\right)=0.$
The total number of root(s) of the equation $f\left(x\right)=g\left(x\right)$ is/are

1. $1$
2. $2$
3. $0$
4. $\infty$

Q. The equation of the curve passing through the origin if the middle point of the segment of its normal form any point of the curve to the x-axis, lies on the parabola $2y^2=x$

1. $y^2=2x+e^{-2x}$
2. $y^2=x+1-e^{-4x}$
3. $y^2=2x+1+e^{2x}$
4. $y^2=2x+1-e^{2x}$

Q. While doing an experiment in chemistry lab, Komal found that a substance loses its moisture at a rate proportional to the moisture content. If the same substance loses half of its moisture during the first hour, when will it lose 99% ?

1. In 50
2. $\frac{ln50}{ln2}$
3. In 25
4. $\frac{ln100}{ln2}$

Q. The equation of a curve passing through $(1,0)$ for which the product of the abscissa of a point $P$ and the intercept made by a normal at $P$ on the $x$-axis equals twice the square of the distance of the point $P$ from the origin $O$, is

1. $x^2+y^2=x^4$
2. $x^2+y^2=2x^4$
3. $x^2+y^2=4x^4$
4. $x^2+y^2=8x^4$

Q. Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth $y$, where the constant of proportionality $k>0$ depends on the acceleration due to gravity and the geometry of the hole. If $t$ is measured in minutes and $k=\frac{1}{15}$, then the time to drain the tank if the water is $4\text{m}$ deep to start with is

1. $30\text{min}$
2. $45\text{min}$
3. $60\text{min}$
4. $80\text{min}$