anonymous
  • anonymous
what is the slope of the tangent line to the curve (x^3 + y)^2 -2a^2xy = b^2 at the point P(0,1) when a=(2)^(1/2) and b = 1
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
anonymous
  • anonymous
I'm getting DNE but its wrong
Jhannybean
  • Jhannybean
For this problem can you input the values for a and b, then find the derivative? Or does the inputting of values happen after differentiating..

Looking for something else?

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

More answers

Jhannybean
  • Jhannybean
It looks really ugly if you differentiate it along with a and b :\
Jhannybean
  • Jhannybean
I would use implicit differentiation to find the derivative.
anonymous
  • anonymous
i really have no idea , iv imputed the values for a and be and then solved for y and derived that , but that seems to be wrong
Jhannybean
  • Jhannybean
Yeah it is hahahahaha
Jhannybean
  • Jhannybean
Because that's basically stating the tangent line is found at a point before even being found.
anonymous
  • anonymous
you only need to find f'(x,y)
anonymous
  • anonymous
but before u start it says when a=(2)^(1/2) and b = 1 , a and b seems like small values and you can sub them why it mention them in ur question ? seems like there is a formula in ur book or something
IrishBoy123
  • IrishBoy123
\[\vec n = <2(x^3+y)3x^2 - 2a^2y, 2(x^3+y)-2a^2x>\] \[= <6x^2(x^3+y) - 2a^2y, 2(x^3+y)-2a^2x>\] x = 0 \[= <-2a^2y,2>\] \[y = 1, a = \sqrt{2}\] \[\vec n = <-4,2,>\]
freckles
  • freckles
don't solve for y just find the derivative of both sides w.r.t. x a and b are constants you can input those values in later or not once you have differentiated both sides you can input the values for a and b if you haven't already and then also input 0 for x and 1 for y then solve for y' to find the slope.
anonymous
  • anonymous
see thats what i'm talking about :D like @irishboy123
IrishBoy123
  • IrishBoy123
|dw:1441778557798:dw|
anonymous
  • anonymous
just noticed the name of the chapter is implicit differention
freckles
  • freckles
why is there y'=that mess=0?
freckles
  • freckles
wait I just had a problem with y'= part
anonymous
  • anonymous
well the question said (tangent line) so it's line equation u can do it in both ways with irish normal vector or with slope and a point
freckles
  • freckles
\[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }\]
Jhannybean
  • Jhannybean
Thanks.
freckles
  • freckles
I just didn't agree with the y'=that stuff=0
freckles
  • freckles
because saying y'=0 is saying you already know the slope to be 0
freckles
  • freckles
\[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }\] so copying this @Jdosio just enter in 0 for x and 1 for y simplify and solve for y'
anonymous
  • anonymous
okay
anonymous
  • anonymous
a^2 ?
freckles
  • freckles
yes and a was?
anonymous
  • anonymous
a = 2^(1/2) so final answer is 2
freckles
  • freckles
right but hey listen do you understand how she got that equation?
anonymous
  • anonymous
not a single clue ^_^
freckles
  • freckles
she used chain rule and product rule (there are some other small potato rules she used)
freckles
  • freckles
\[\frac{d}{dx}(x^3+y)^2 \] for this you should use chain rule
Jhannybean
  • Jhannybean
\[ (x^3 + y)^2 -2a^2xy = b^2\]\[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \]\[6x^5+2x^3y'+6x^3y+2yy'-2a^2xy'-2a^2y=0\]\[y'(2x^3+2y-2a^2x)=-6x^5 -6x^3y+2a^2y\]\[y'=...\]
freckles
  • freckles
\[\frac{d}{dx}(x^3+y)^2 =2(x^3+y) \cdot (x^3+y)'\] derivative of outside times the derivative of inside
freckles
  • freckles
\[\frac{d}{dx}(x^3+y)=3x^2+y' \\ \text{ so } \frac{d}{dx}(x^3+y)^2=2(x^3+y) \cdot (3x^2+y')\]
Jhannybean
  • Jhannybean
chain rule: \(f(g(x)) = f'(x) \cdot g'(x)\)
anonymous
  • anonymous
oh okay i see i see
anonymous
  • anonymous
i was just rather lost on the second part of the subtraction where you used the product rule , im familiar with it i just wasn't entirely sure how to go about the a but i see what you did
freckles
  • freckles
the other term you can use constant multiple rule and product rule \[\frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy)\]
anonymous
  • anonymous
yea x' y + x y'
Jhannybean
  • Jhannybean
You're trying to keep in mind that taking the derivative of \(y\) will give you \(\dfrac{dy}{dx}\) and that you are taking the derivative of \(x\) as well.
freckles
  • freckles
that is correct but x'=?
freckles
  • freckles
or dx/dx=?
anonymous
  • anonymous
wait , are you asking about the example or the actual question?
freckles
  • freckles
well you said \[\frac{d}{dx}(xy)=\frac{dx}{dx} y+x \frac{dy}{dx}\] and I was asking you to simplify dx/dx
anonymous
  • anonymous
well 1 right?
freckles
  • freckles
yep
freckles
  • freckles
\frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy) right \[\frac{d}{dx}(xy)=y+xy' \\ \text{ so } \frac{d}{dx}-2a^2xy=-2a^2(y+xy') =-2a^2y-2a^2xy'\]
freckles
  • freckles
and of course the other side of the equation is b^2 so when differentiating a constant you get 0
freckles
  • freckles
@Jdosio do you have any questions on this?
anonymous
  • anonymous
not on this question specifically , if i come up with anything else i missed ill ask you guys if your still around. thanks for all your help once again.
Jhannybean
  • Jhannybean
No problem.

Looking for something else?

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