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anonymous
 one year ago
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
anonymous
 one year ago
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @Robert136 @nincompoop

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm getting DNE but its wrong

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1For this problem can you input the values for a and b, then find the derivative? Or does the inputting of values happen after differentiating..

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1It looks really ugly if you differentiate it along with a and b :\

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1I would use implicit differentiation to find the derivative.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.1Yeah it is hahahahaha

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1Because that's basically stating the tangent line is found at a point before even being found.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you only need to find f'(x,y)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\vec n = <2(x^3+y)3x^2  2a^2y, 2(x^3+y)2a^2x>\] \[= <6x^2(x^3+y)  2a^2y, 2(x^3+y)2a^2x>\] x = 0 \[= <2a^2y,2>\] \[y = 1, a = \sqrt{2}\] \[\vec n = <4,2,>\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1don'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
 one year ago
Best ResponseYou've already chosen the best response.0see thats what i'm talking about :D like @irishboy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441778557798:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just noticed the name of the chapter is implicit differention

freckles
 one year ago
Best ResponseYou've already chosen the best response.1why is there y'=that mess=0?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1wait I just had a problem with y'= part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.1\[2(x^3+y)(3x^2+y')(2a^2xy'+2a^2y)=0 \text{ was right }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I just didn't agree with the y'=that stuff=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1because saying y'=0 is saying you already know the slope to be 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[2(x^3+y)(3x^2+y')(2a^2xy'+2a^2y)=0 \text{ was right }\] so copying this @Jdosio just enter in 0 for x and 1 for y simplify and solve for y'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a = 2^(1/2) so final answer is 2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1right but hey listen do you understand how she got that equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not a single clue ^_^

freckles
 one year ago
Best ResponseYou've already chosen the best response.1she used chain rule and product rule (there are some other small potato rules she used)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx}(x^3+y)^2 \] for this you should use chain rule

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[ (x^3 + y)^2 2a^2xy = b^2\]\[2(x^3+y)(3x^2+y')(2a^2xy'+2a^2y)=0 \]\[6x^5+2x^3y'+6x^3y+2yy'2a^2xy'2a^2y=0\]\[y'(2x^3+2y2a^2x)=6x^5 6x^3y+2a^2y\]\[y'=...\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx}(x^3+y)^2 =2(x^3+y) \cdot (x^3+y)'\] derivative of outside times the derivative of inside

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx}(x^3+y)=3x^2+y' \\ \text{ so } \frac{d}{dx}(x^3+y)^2=2(x^3+y) \cdot (3x^2+y')\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1chain rule: \(f(g(x)) = f'(x) \cdot g'(x)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.1the other term you can use constant multiple rule and product rule \[\frac{d}{dx}2a^2xy \\ 2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy)\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1You'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
 one year ago
Best ResponseYou've already chosen the best response.1that is correct but x'=?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait , are you asking about the example or the actual question?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well you said \[\frac{d}{dx}(xy)=\frac{dx}{dx} y+x \frac{dy}{dx}\] and I was asking you to simplify dx/dx

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\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 \[\frac{d}{dx}(xy)=y+xy' \\ \text{ so } \frac{d}{dx}2a^2xy=2a^2(y+xy') =2a^2y2a^2xy'\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and of course the other side of the equation is b^2 so when differentiating a constant you get 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@Jdosio do you have any questions on this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not 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.
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