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We have that The height changing rate when the length of the base is 2 cm
[tex]dh/dt=-7.2cm/s[/tex]
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
[tex]r^2=h^2+x^2[/tex]
With r=3
Therefore
dx/dt=8cm/s
Hence
[tex]2rdr/dt=2hdh/dt+2xdx/dt\\\\\dh/dt=\frac{-x}{r^2-x^2}dx/dt[/tex]
We have
[tex]dh/dt=-2/\sqrt{5}(8)\\\\dh/dt=-7.2cm/s[/tex]
Therefore
The height changing rate when the length of the base is 2 cm
[tex]dh/dt=-7.2cm/s[/tex]
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