Implicitly Deifned Parametrization

In summary, implicit differentiation can be done by differentiation of both equations w.r.t.t. dummy variables, but in order to find the slope one must find x and y in the original equation.
  • #1
GregA
210
0
Implicitly Defined Parametrization

I'm having difficulties with the following question, and having checked through my working several times I just can't find a problem...problem is, so far in the book implicit and parametric differentiation have been covered independently of each other and this question has just been thrown into the mix.

Find the slope when t = 0:
[tex] x + 2x^{\frac {3}{2}} = t^2 + t[/tex], [tex] y \sqrt{t+1} + 2t \sqrt{y} = 4 [/tex]
I can't see any other way to work with this other than to differentiate both equations w.r.t.t, find dy/dx by dividing dy/dt by dx/dt, and then plug in my numbers at the end.

dx/dt: [tex] \frac{dx}{dt} +3 \sqrt{x} \frac{dx}{dt} = 2t + 1[/tex]

[tex] \frac{dx}{dt} =\frac{2t + 1}{3 \sqrt{x}+1}[/tex]

dy/dt: [tex] \sqrt{t+1} \frac{dy}{dt} + \frac {y}{2\sqrt{t+1}} + \frac{t}{\sqrt{y}}\frac{dy}{dt}+ 2\sqrt{y} = 0[/tex]
just going to assign dummy variables:[tex] a = \sqrt{y}, b = \sqrt{t+1} [/tex] whilst I rearrange things...
[tex] \frac{dy}{dt}(b + \frac{t}{a}) = -(\frac {y}{2b} +2a)[/tex]

[tex] \frac{dy}{dt}(\frac{ba +t}{a}) = -(\frac {y +4ab}{2b})[/tex]

[tex] \frac{dy}{dt} = -(\frac {a(y +4ab)}{2b(ba+t)}) = -(\frac {\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1})}{2\sqrt{t+1}(\sqrt{y}\sqrt{t+1}+t)})[/tex]

dy/dx =[tex] -\frac {(3 \sqrt{x}+1)(\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1}))} {(2t + 1)(2\sqrt{t+1}(\sqrt{y}\sqrt{t+1}+t))}[/tex]

By using t = 0 and finding x and y in the original equations I get x = 1/4 and y = 4
[tex] -\frac{(\frac{3}{2}+1)(8 + 16)}{4} = -15 [/tex]

The answer I'm looking for however is -6. Is there some screw up with my working somewhere or am I using the wrong method?
I cannot graph any similar (but simpler) curves with Maxima to find the correct method neither because there doesn't seem to be a way to graph implicit expressions...nor will it differentiate one
 
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  • #2
GregA said:
I'm having difficulties with the following question, and having checked through my working several times I just can't find a problem...problem is, so far in the book implicit and parametric differentiation have been covered independently of each other and this question has just been thrown into the mix.

Find the slope when t = 0:
[tex] x + 2x^{\frac {3}{2}} = t^2 + t[/tex], [tex] y \sqrt{t+1} + 2t \sqrt{y} = 4 [/tex]
I can't see any other way to work with this other than to differentiate both equations w.r.t.t, find dy/dx by dividing dy/dt by dx/dt, and then plug in my numbers at the end.

EDIT: Indeed it gives -6 with x=0, y=4

dx/dt: [tex] \frac{dx}{dt} +3 \sqrt{x} \frac{dx}{dt} = 2t + 1[/tex]

[tex] \frac{dx}{dt} =\frac{2t + 1}{3 \sqrt{x}+1}[/tex]

dy/dt: [tex] \sqrt{t+1} \frac{dy}{dt} + \frac {y}{2\sqrt{t+1}} + \frac{t}{\sqrt{y}}\frac{dy}{dt}+ 2\sqrt{y} = 0[/tex]
just going to assign dummy variables:[tex] a = \sqrt{y}, b = \sqrt{t+1} [/tex] whilst I rearrange things...
[tex] \frac{dy}{dt}(b + \frac{t}{a}) = -(\frac {y}{2b} +2a)[/tex]

[tex] \frac{dy}{dt}(\frac{ba +t}{a}) = -(\frac {y +4ab}{2b})[/tex]

[tex] \frac{dy}{dt} = -(\frac {a(y +4ab)}{2b(ba+t)}) = -(\frac {\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1})}{2\sqrt{t+1}(\sqrt{y}\sqrt{t+1}+t)})[/tex]

dy/dx =[tex] -\frac {(3 \sqrt{x}+1)(\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1}))} {(2t + 1)(2\sqrt{t+1})\sqrt{y}\sqrt{t+1}+t))}[/tex]

By using t = 0 and finding x and y in the original equations I get x = 1/4 and y = 4
[tex] -\frac{(\frac{3}{2}+1)(8 + 16)}{4} = -15 [/tex]

The answer I'm looking for however is -6. Is there some screw up with my working somewhere or am I using the wrong method?
I cannot graph any similar (but simpler) curves with Maxima to find the correct method neither because there doesn't seem to be a way to graph implicit expressions...nor will it differentiate one
I did not check all the steps (the parenthesis don't match in the denominator in yoru final dy/dx) but how did you get x=1/4? When t =0, x=0 !
 
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  • #3
Fixed the parenthisis :smile:
By setting t =0 in:[tex] x + 2x^{\frac {3}{2}} = t^2 + t[/tex]
I get :[tex] x + 2x^{\frac {3}{2}} = 0[/tex]

[tex] x^{\frac {1}{2}} = -\frac{1}{2}[/tex]
x = 1/4

That response has just made me see how -6 is possible tho :smile:
 
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  • #4
GregA said:
Fixed the parenthisis :smile:
By setting t =0 in:[tex] x + 2x^{\frac {3}{2}} = t^2 + t[/tex]
I get :[tex] x + 2x^{\frac {3}{2}} = 0[/tex]

[tex] x^{\frac {1}{2}} = -\frac{1}{2}[/tex]
x = 1/4

You are right in that there are two solutions. The other solution (x=0) gives indeed -6.
It`s all a question of which sign to take when calculating a square roo. For your answer to work one must use the convention that one must take the negative sign for a square root. I was assuming that the convention was to take the positive root.
 
  • #5
Thankyou very much Nrged for putting my mind at ease and pointing out a solution I kept overlooking...what I did *lost* a solution! :smile:
 
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  • #6
Actually, no. [itex]x^{\frac{1}{2}}= -\frac{1}{2}[/itex] has no solution since the 1/2 power of any real number is non-negative. The only root os [itex]x+ 2x^{\frac{3}{2}}= 0 is 0.
 
  • #7
Thanks for pointing out where I am wrong HallsofIvy :smile: I won't make this same mistake twice.
 
  • #8
HallsofIvy said:
Actually, no. [itex]x^{\frac{1}{2}}= -\frac{1}{2}[/itex] has no solution since the 1/2 power of any real number is non-negative. The only root os [itex]x+ 2x^{\frac{3}{2}}= 0 is 0.
So if we have x^2 = 4, the solution is only x=2 and cannot be x=-2? (where x is a real) ?
 
  • #9
no because x^2 = 4 does give two roots. +2 and -2.

-2*-2 = 4

2*2 = 4
 
  • #10
mathmike said:
no because x^2 = 4 does give two roots. +2 and -2.

-2*-2 = 4

2*2 = 4
I know, but HallsofIvy seems to imply the contrary which kind of surprised me . I'd like to understand what he meant (he is very knowledgeable so I must be missing his point entirely)
 
  • #11
HallsofIvy said:
Actually, no. [itex]x^{\frac{1}{2}}= -\frac{1}{2}[/itex] has no solution since the 1/2 power of any real number is non-negative. The only root os [itex]x+ 2x^{\frac{3}{2}}= 0 is 0.

I am still confused by that statement: since the 1/2 power of any real number is non-negative.
 
  • #12
The equation x2= a has two solutions. They can be written [itex]\sqrt{a}[/itex] and [itex]-\sqrt{a}[/itex]. The reason why we need to write "+" and "-" is because [itex]\sqrt{a}[/itex] is defined to be the positive root! There is a difference between "the roots of the equation x2= a" and "the square root of a". The equation has two roots while the square root function, [itex]\sqrt{a}[/itex], is defined as the positive solution to x2= a.
a1/2 is also [ b]defined[/b] as [itex]\sqrt{a}[/itex], the positive real root of x2[/sub]= a.
 
  • #13
HallsofIvy said:
The equation x2= a has two solutions. They can be written [itex]\sqrt{a}[/itex] and [itex]-\sqrt{a}[/itex]. The reason why we need to write "+" and "-" is because [itex]\sqrt{a}[/itex] is defined to be the positive root! There is a difference between "the roots of the equation x2= a" and "the square root of a". The equation has two roots while the square root function, [itex]\sqrt{a}[/itex], is defined as the positive solution to x2= a.
a1/2 is also [ b]defined[/b] as [itex]\sqrt{a}[/itex], the positive real root of x2[/sub]= a.

Ok, that makes sense. Thank you.
 

1. What is Implicitly Defined Parametrization?

Implicitly Defined Parametrization is a mathematical concept used to represent a curve or surface in terms of one or more variables. It involves defining the curve or surface implicitly, rather than explicitly, using a set of equations.

2. How is Implicitly Defined Parametrization different from Explicitly Defined Parametrization?

In Explicitly Defined Parametrization, the curve or surface is represented in terms of a single parameter. However, in Implicitly Defined Parametrization, the curve or surface is represented in terms of multiple parameters, making it a more generalized form.

3. What are the advantages of using Implicitly Defined Parametrization?

Implicitly Defined Parametrization allows for a more flexible and versatile representation of curves and surfaces, as it can handle complex shapes and curves that may not be easily defined explicitly. It also allows for the representation of curves or surfaces that cannot be represented explicitly.

4. How is Implicitly Defined Parametrization used in scientific research?

Implicitly Defined Parametrization is commonly used in fields such as physics, engineering, and computer graphics to model and analyze complex shapes and surfaces. It is also used in numerical analyses and simulations to solve differential equations.

5. Are there any limitations to using Implicitly Defined Parametrization?

While Implicitly Defined Parametrization offers many advantages, it can also be more computationally intensive and complex compared to Explicitly Defined Parametrization. It may also be more challenging to visualize and interpret the results, as the equations involved may not have a direct geometric interpretation.

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