Respuesta :
Answer:
[tex]y[/tex] is decreasing by 10 units per second.
Step-by-step explanation:
Given information:
The particle moves on the hyperbola [tex]xy=15[/tex]
Time [tex]t\geq 0[/tex]
Now at a certain instant , [tex]x=3[/tex] and [tex]\frac{dx}{dt} = 6[/tex]
Now, Differentiating [tex]xy=15[/tex] with respect to [tex]t[/tex]
We get:
[tex]x.\frac{dy}{dt} +y.\frac{dx}{dt} =0[/tex]
Now substitute the values in above equation:
[tex]3\frac{dy}{dt}+ 5 \times 6=0[/tex]
[tex]3 \frac{dy}{dt}=-30\\\frac{dy}{dt} =-10[/tex]
The negative sign indicates that [tex]y[/tex] is decreasing by 10 units per second.
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The particle follows hyperbolic path. The value of [tex]y[/tex] is decreasing at the rate of [tex]10[/tex] units per seconds.
Given: A particle moves on the hyperbola [tex]xy=15[/tex] for time [tex]t\geq0[/tex] seconds. At a certain instant, [tex]x=3[/tex] and [tex]dx/dt=6[/tex].
According to question,
hyperbola curve is [tex]xy=15[/tex].
Differentiating the curve w.r.t [tex]t[/tex] we get:
[tex]x\cdot \frac{dy}{dt}+y\cdot\frac{dx}{dt}=0[/tex]
Now substituting the values are [tex]x=3[/tex] and [tex]dx/dt=6[/tex].
[tex]3\cdot \frac{dy}{dt}+5\times6=0[/tex]
[tex]\frac{dy}{dt}=\frac{-30}{3}\\\frac{dy}{dt}=-10[/tex]
Therefore, [tex]y[/tex] is decreasing [tex]10[/tex] units per seconds.
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