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Josee
The height of a triangle is increasing at a rate of 5 cm/min while its area is increasing at a rate of 4 sq. cms/min. At what speed is the base of the triangle changing when the height of the triangle is 6 cms and its area is 24 sq. cms?
\[\text{Dt}[b]=-\frac{16}{3}\text{cm}/\min \]
\[A=\frac{1}{2}BH.\]\[\frac{\partial H}{\partial t}=5,\]\[\frac{\partial A}{\partial t}=4.\]\[B=2\frac{A}{H},\]\[\frac{\partial B}{\partial t}=2\frac{H\frac{\partial A}{\partial t}-\frac{\partial H}{\partial t}A}{H^2}=2\frac{(6)(4)-(5)(24)}{(6)^2}=2\frac{24-120}{36}=-\frac{16}{3},\]as robtobey said.
\[\text{area}\text{= }\frac{b h}{2}\]\[b=\frac{2 \text{area}}{h} \]Take the total derivative of the expression above\[\text{Dt}[b]=\frac{2 \text{Dt}[\text{area}]}{h}-\frac{2 \text{area} \text{Dt}[h]}{h^2}\]\[\{ h\to 6,\text{area}\to 24,\text{Dt}[\text{area}]\to 4, \text{Dt}[h]\to 5\} \]\[\text{Dt}[b]==\frac{2*4}{6}-\frac{2*24*5}{6^2} \]\[\text{Dt}[b]=-\frac{16}{3}\]