Mathematics: [calculus] i this guy out here on on a problem with related rates. does anyone want to check my work? the question can be found here: can someone ?

Here is the work and answers to your question hope this helps....

[calculus] i this guy out here on on a problem with

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aj2869

31.07.2019

See diagram.in your answer, you incorrectly assumed that $dh/dt = 0$ .the "area of the triangle formed by the wall, the ground, and the ladder" is represented in gray. $dh/dt$ is changing. as you pull the ladder away, distance from the top of the ladder to the bottom ( $h$ ) decreases.the rate at which "h" changes at that instant can be found using the pythagorean theorem. $\begin{aligned} h^2 + x^2 & = 15\\ 2h \frac{dh}{dt} + 2x \frac{dx}{dt} & = 0 \\ h \frac{dh}{dt} + x \frac{dx}{dt} & = 0 \\ h \frac{dh}{dt} & = - x \frac{dx}{dt} \\ \frac{dh}{dt} & = -\frac{x}{h} \cdot \frac{dx}{dt} \end{aligned}$ side note: division is just multiplication (reciprocal'd) $\begin{aligned} \dfrac{a\cdot b}{c} & = (a\cdot b) \div c \\ & = a\cdot b \div c \\ & = a\cdot b \cdot \dfrac{1}{c} \end{aligned}$ multiplication order does not matter. $\begin{aligned} \dfrac{a\cdot b}{c} & = a\cdot b \cdot \dfrac{1}{c} \\ & = a\cdot \dfrac{b}{c} \\ & = \frac{a}{c} \cdot b \end{aligned}$ so $\begin{aligned} -\frac{x}{h} \cdot \frac{dx}{dt} & = \frac{-x \cdot \frac{dx}{dt}}{h}\\ & = -x \cdot \frac{\ \frac{dx}{dt}\ }{h} \end{aligned}$ but need to be more careful when dealing with sums, e.g. $\begin{aligned} (x+y) \div c & = (x+y) \cdot \dfrac{1}{c} \\ & = \dfrac{x+y}{c} \\ & = \dfrac{x}{c} + \dfrac{y}{c} \end{aligned}$

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[calculus] i this guy out here on on a problem with related rates. does anyone want to check my work? the question can be found here: can someone ?

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