# Thread: Further Differentiation rates of changes

1. ## Further Differentiation rates of changes

a horse trough has a triangular cross section of height 25 cm and base 30 cm, and 2m long. A horse drinking steadily and when the water level is 5 cm below the top it is being lowered at the rate of 1 cm/ min. find the rate of consumption in litres per minute

My attempt:
v = $\displaystyle \frac{1}{2}bh \times l$
thus v =
$\displaystyle b = \frac{6h}{5}$
$\displaystyle v = \frac{6h^2}{5}$
$\displaystyle \frac{\mathrm{d}v }{\mathrm{d} h} =\frac{12h}{5}$
$\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= \frac{12h}{5} \times \frac{\mathrm{d}h }{\mathrm{d} t}$
since the water level is lowered by 5 cm thus h = 20 cm
thus;
$\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= 48 \times \frac{\mathrm{d}h }{\mathrm{d} t}$

$\displaystyle \frac{\mathrm{d}c }{\mathrm{d} t}=\frac{\mathrm{d}c }{\mathrm{d} h} \times \frac{\mathrm{d}h }{\mathrm{d} t}$
im stuck here

thank you

2. ## Re: Further Differentiation rates of changes

Originally Posted by bigmansouf
a horse trough has a triangular cross section of height 25 cm and base 30 cm, and 2m long. A horse drinking steadily and when the water level is 5 cm below the top it is being lowered at the rate of 1 cm/ min. find the rate of consumption in litres per minute

My attempt:
v = $\displaystyle \frac{1}{2}bh \times l$
thus v =
$\displaystyle b = \frac{6h}{5}$
$\displaystyle v = \frac{6h^2}{5}$ Watch your units … length = 2m=200cm … v =120h^2
$\displaystyle \frac{\mathrm{d}v }{\mathrm{d} h} =\frac{12h}{5}$
$\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= \frac{12h}{5} \times \frac{\mathrm{d}h }{\mathrm{d} t}$
since the water level is lowered by 5 cm thus h = 20 cm
thus;
$\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= 48 \times \frac{\mathrm{d}h }{\mathrm{d} t}$

$\displaystyle \frac{\mathrm{d}c }{\mathrm{d} t}=\frac{\mathrm{d}c }{\mathrm{d} h} \times \frac{\mathrm{d}h }{\mathrm{d} t}$
im stuck here