For equally inclined, consider \(\frac{d y}{d x}=\pm 1\)  (4,4)

Given that L represents the number of litres of water in the pool, t seconds after the pool has been plugged off todrain, then 
\(L=200(10-t)^{2}\) 
ஃ Rate at which the water is running out
\(\frac{d L}{d t}=-200 \cdot 2(10-t) \cdot(-1)=400(10-t)\) 
[here, we take negative sign, because the water is decreasing]
Rate at which the water is running out at the end of 5s
= 400(10 - 5)= 2000 L /s = Final rate
Initial rate \((\text { at } t=0)=\left(\frac{d L}{d t}\right)_{t=0}=4000 \mathrm{~L} / \mathrm{s}\) 
\( \therefore \text { Average rate during } 5 \mathrm{~s} =\frac{\text { Initial rate }+\text { Final rate }}{2} \)
\(=\frac{4000+2000}{2}=3000 \mathrm{~L} / \mathrm{s} \) 
Hence, the water flows out during the first 5s withaverage rate of 3000 L / s

Maximum slope m=-3+6+9=12

\(tan^-1{3\over 4}\)

4y+2x+\(\pi=0\)

Find point of intersection on two curves and show that slopes of tangents to the two curves at point of intersection is same.

\({x\over a}+{y\over b}=1\)

The area A₁ of square of side x is given by A₁= x²
and area A₂ of square of side y is given by
\(A_{2}=y^{2}=\left(x-x^{2}\right)^{2} \)
\(\frac{d A_{1}}{d x} =2 x, \frac{d A_{2}}{d x}=2\left(x-x^{2}\right)(1-2 x) \)
\(\frac{d A_{2}}{d A_{1}} =\frac{d A_{2}}{d x}= \frac{d A_{1}}{d x}=\frac{2\left(x-x^{2}\right)(1-2 x)}{2 x} \)
\(=(1-x)(1-2 x)=1-3 x+2 x^{2} \)