anonymous
  • anonymous
can someone explain to me Implicit Differentiation: if y^3 = 25x^2, determine dx/dt when x = 5 and dy/dt = 1
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
isn't y=2.92402?
anonymous
  • anonymous
i did this on the calculator: sqrt (625)^(1/2)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
(1/3)**
anonymous
  • anonymous
@satellite73
agent0smith
  • agent0smith
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 \[\Large 3 y^2 *\frac{ dy }{ dt }\]
anonymous
  • anonymous
so I can't replace the dy/dt by 1 since it's given?
anonymous
  • anonymous
@agent0smith
agent0smith
  • agent0smith
Well yes but i was seeing if you understood the differentiation... don't worry about plugging in numbers till later.
anonymous
  • anonymous
and 3y^2 is already the derivative, why do we need to use the chain rule?
anonymous
  • anonymous
@agent0smith
agent0smith
  • agent0smith
Because you have to multiply by the derivative of y with respect to t.
agent0smith
  • agent0smith
3y^2 is NOT the derivative of y^3, UNLESS you're just differentiating with respect to y.
anonymous
  • anonymous
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
  • agent0smith
It's the same with dy/dx... the derivative of y with respect to x is not just 1... it's 1*dy/dx
anonymous
  • anonymous
ok... :/ it's a little confusing..
ranga
  • ranga
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
  • agent0smith
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.
anonymous
  • anonymous
why can't we plug in the numbers right away if it's given?
agent0smith
  • agent0smith
You can... you're probably just better off not doing it until you really know what you're doing.
anonymous
  • anonymous
right i understand that... @ranga
anonymous
  • anonymous
well i was taught to plug in.. that's why.
anonymous
  • anonymous
that's the confusing part.
agent0smith
  • agent0smith
Fair enough... but you can't plug in until you have this step: 3y^2*dy/dt = 50x*dx/dt
anonymous
  • anonymous
exactly what i have right now.
anonymous
  • anonymous
so dy/dt is 1...
anonymous
  • anonymous
in the steps above, why is this: 3(8.55)^3*1 = 50*5*dx/dt?
anonymous
  • anonymous
y=8.55? y is suppose to be 2.92402.
anonymous
  • anonymous
and it's raised to the 3rd, and why not 2?
agent0smith
  • agent0smith
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
anonymous
  • anonymous
so y=2.92402 is correct?
agent0smith
  • agent0smith
y^3 = 625 y = cube root 625 = 8.55
anonymous
  • anonymous
how?
anonymous
  • anonymous
i did this in the calculator: sort(625)^(1/3)
anonymous
  • anonymous
sqrt*
agent0smith
  • agent0smith
And you'll get 8.55... 2.94 cubed is close to 3^3 which is 27. Not 625.
anonymous
  • anonymous
i swear im getting 2.92402
anonymous
  • anonymous
i plugged it in exactly like iwrote it.
agent0smith
  • agent0smith
sort(625)^(1/3) why are you taking the square root of a cubed root...
agent0smith
  • agent0smith
y^3 = 625 y = 625^(1/3)
anonymous
  • anonymous
OHHH
anonymous
  • anonymous
i thought the sqrt could do cube root.
anonymous
  • anonymous
if we did the exponent also.
agent0smith
  • agent0smith
That would mean ((625^(1/3))^(1/2) = 625^(1/6)
anonymous
  • anonymous
thank you! i got it !

Looking for something else?

Not the answer you are looking for? Search for more explanations.