We have a bug moving alone the curve y = 4 - x^2/16. The distance is measured in feet. When the bug passes through point (4,3) its y-coordinate is increasing at 20 ft/sec so at what rate is its x-coordinate changing?
:/ sorry not good with this kind of stuff, i could call for help tho @taylorannbs @makayla_finkbeiner_ @Bookworm14 @aprilann143
wow thanks man
no prob
you're given : \(\dfrac{dy}{dt} = 20\)
you need to find \(\dfrac{dx}{dt}\)
differentiate the given function with.respect.to \(t\)
wait this is lagging
yeah, just differentiate ur function, plugin given values, solve for dx/dt
so that is t/x + 1/8t^2 with implicit diff right?
\(y = 4 - \dfrac{x^2}{16}\)
Is that ur given curve ?
ok so i need to find dx/dt of that right?
I believe so, that would give me -t^2(x) right?
what do u mean u believe so ? you're not given the question/equation in correct format ?
whats wid two `//` in the question ? `y = 4 - x^2//16` what kind of operator is a `//` ?
ohhhh that was an accident
lol okay, take a snapshot of the question and attach if psble
i changed it and made it right
okay :)
\(y = 4 - \dfrac{x^2}{16}\) differentiate both sides with.respect.to \(t\) : \(\dfrac{dy}{dt} = 0 - \dfrac{1}{16}(2x)\dfrac{dx}{dt} \)
^^simplify
one second
lol i'm horrible with this but I got x/2t + 4t/x
wait no dy/dt = -32x dx/dt
could u show me how u got that ?
careful, 2 is in numerator and 16 is in denominator
\(\dfrac{dy}{dt} = 0 - \dfrac{1}{16}(2x)\dfrac{dx}{dt} \) simplifies to \(\dfrac{dy}{dt} = - \dfrac{x}{8}\dfrac{dx}{dt} \)
oh well i got the other one by simplifying 0-1/16(2x)
plugin the given values : \(x = 4\) \(\dfrac{dy}{dt} = 20\) and solve \(\dfrac{dx}{dt}\)
\(\dfrac{dy}{dt} = - \dfrac{x}{8}\dfrac{dx}{dt} \) \(20 = - \dfrac{4}{8}\dfrac{dx}{dt} \)
solve \(\dfrac{dx}{dt}\)
it's not -40 is it?
-40 is right !
no way
that's final?
thats final.
well thanks a lot i mean i got a study guide and i need the help to get ready for this final
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