mathcalculus
can someone explain to me Implicit Differentiation:
if y^3 = 25x^2, determine dx/dt when x = 5 and dy/dt = 1
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mathcalculus
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these are the steps but i don't understand them :/
3y^2*dy/dt = 50x*dx/dt
When x = 5, y^3 = 25*5^2
y^3 = 625
y = cube root 625 = 8.55
3(8.55)^3*1 = 50*5*dx/dt
1875 = 250dx/dt
7.5 = dx/dt
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isn't y=2.92402?
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i did this on the calculator: sqrt (625)^(1/2)
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(1/3)**
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@satellite73
agent0smith
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Let's start with this part...
\[\Large \frac{ d }{ dt } y^3\]use the chain rule to differentiate it... so bring down the exponent, reduce the exponent by one, then multiply by the derivative of y w. respect to t \[\Large 3 y^2 *\frac{ dy }{ dt }\]
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so I can't replace the dy/dt by 1 since it's given?
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@agent0smith
agent0smith
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Well yes but i was seeing if you understood the differentiation... don't worry about plugging in numbers till later.
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and 3y^2 is already the derivative, why do we need to use the chain rule?
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@agent0smith
agent0smith
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Because you have to multiply by the derivative of y with respect to t.
agent0smith
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3y^2 is NOT the derivative of y^3, UNLESS you're just differentiating with respect to y.
tpmys
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normally we have y dependent on x
here y, and x both are dependent on some t, so we dont know what the function is
so instead of y^3 >>> 3y^2 we have the chain rule y^3 >>> 3y^2 * y'
agent0smith
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It's the same with dy/dx... the derivative of y with respect to x is not just 1... it's 1*dy/dx
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ok... :/ it's a little confusing..
ranga
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agent0smith explained it nicely:
\[\frac{ d }{ dy }(y ^{3}) =3y ^{2}\]
But
\[\frac{ d }{ dt }y ^{3} = \frac{ d }{ dy }y ^{3}\frac{ dy }{ dt } = 3y ^{2}\frac{ dy }{ dt }\]
agent0smith
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It's the same thing with the chain rule \[\huge \frac{ d }{ dx } f(x)^n = n*f(x)^{n-1} * f \prime (x)\]
you have to always multiply by the derivative at the end.
mathcalculus
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why can't we plug in the numbers right away if it's given?
agent0smith
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You can... you're probably just better off not doing it until you really know what you're doing.
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right i understand that... @ranga
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well i was taught to plug in.. that's why.
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that's the confusing part.
agent0smith
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Fair enough... but you can't plug in until you have this step: 3y^2*dy/dt = 50x*dx/dt
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exactly what i have right now.
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so dy/dt is 1...
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in the steps above, why is this: 3(8.55)^3*1 = 50*5*dx/dt?
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y=8.55? y is suppose to be 2.92402.
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and it's raised to the 3rd, and why not 2?
agent0smith
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When x = 5, y^3 = 25*5^2
y^3 = 625
y = cube root 625 = 8.55
i think the next line after that is a mistake and it should by to the power of 2
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so y=2.92402 is correct?
agent0smith
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y^3 = 625
y = cube root 625 = 8.55
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how?
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i did this in the calculator: sort(625)^(1/3)
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sqrt*
agent0smith
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And you'll get 8.55... 2.94 cubed is close to 3^3 which is 27. Not 625.
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i swear im getting 2.92402
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i plugged it in exactly like iwrote it.
agent0smith
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sort(625)^(1/3)
why are you taking the square root of a cubed root...
agent0smith
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y^3 = 625
y = 625^(1/3)
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OHHH
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i thought the sqrt could do cube root.
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if we did the exponent also.
agent0smith
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That would mean ((625^(1/3))^(1/2) = 625^(1/6)
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thank you! i got it !