A ladder $10 \,m$ long rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of $3\, cm/s$. The height of the upper end while it is descending at the rate of $4 \,cm/s$, is
Updated On: Jul 5, 2022
$4\sqrt{3}\,m$
$5\sqrt{3}\,m$
$6\,m$
$8\,m$
Hide Solution
Verified By Collegedunia
The Correct Option isC
Solution and Explanation
Let $AB = x\, m$,
$BC = y\, m$ and $AC = 10\, m$$\therefore x^{2 }+ y^{2}= 100 \quad...\left(i\right)$
On differentiating $\left(i\right)$ w.r.t. $t$, we get
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$\Rightarrow 2x\left(3\right) - 2y\left(4\right)= 0$$\Rightarrow x = \frac{4y}{3}$
On putting this value in $\left(i\right)$, we get
$\frac{16}{9} y^{2} + y^{2} = 100$$\Rightarrow y^{2} = \frac{100 \times 9}{25} = 36$$\Rightarrow y = 6\,m$
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
