Question

Question 1: The diagonal of a rectangle, with base x, remains at a fixed
value of 3cm. If the base is increasing at a rate of 8 cm/s. at what rate is
the height changing when the length of the base is 2 cm?

Answer 1

We have that The height changing rate  when the length of the base is 2 cm

$dh/dt=-7.2cm/s$

From the question we are told

The diagonal of a rectangle, with base x, remains at a fixed value of 3cm. If the base is increasing at a rate of 8 cm/s. at what rate is  the height changing when the length of the base is 2 cm?

Generally the Pythagoras equation for the triangle   is mathematically given

as

$r^2=h^2+x^2$

With r=3

Therefore

dx/dt=8cm/s

Hence

$2rdr/dt=2hdh/dt+2xdx/dt\\\\\dh/dt=\frac{-x}{r^2-x^2}dx/dt$

We  have

$dh/dt=-2/\sqrt{5}(8)\\\\dh/dt=-7.2cm/s$

Therefore

The height changing rate  when the length of the base is 2 cm

$dh/dt=-7.2cm/s$

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