How do you find the value of t for which (dy/dx)=0 ? .... given x= 80t and y=64t -16 t^2

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Answer 1 · 2018-05-14T11:45:11

$t = 2$ $t=2$

First equation:

$x = 80 t$ $x=80t$

Hence,

$t = \frac{x}{80}$ $t=x/80$

Differentiate,

$\frac{d t}{d x} = \frac{1}{80}$ $dt/dx=1/80$

Second equation:

$y = 64 t - 16 t^{2}$ $y=64t-16t^2$

Differentiate

$\frac{d y}{d t} = 64 - 32 t$ $dy/dt=64-32t$

Apply chain rule:

$\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$ $dy/dx=dy/dtxxdt/dx$

Hence,

$\frac{d y}{d x} = (64 - 32 t) (\frac{1}{80})$ $dy/dx=(64-32t)(1/80)$

Simplify,

$\frac{d y}{d x} = \frac{4}{5} - \frac{2}{5} t$ $dy/dx=4/5-2/5t$

When $\frac{d y}{d x} = 0$ $dy/dx=0$ ,

$\frac{4}{5} - \frac{2}{5} t = 0$ $4/5-2/5t=0$

Simplify,

$\frac{2}{5} t = \frac{4}{5}$ $2/5t=4/5$

Solve,

$t = 2$ $t=2$