# Homework Help: Implicitly Deifned Parametrization

1. Apr 23, 2006

### GregA

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:
$$x + 2x^{\frac {3}{2}} = t^2 + t$$, $$y \sqrt{t+1} + 2t \sqrt{y} = 4$$
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: $$\frac{dx}{dt} +3 \sqrt{x} \frac{dx}{dt} = 2t + 1$$

$$\frac{dx}{dt} =\frac{2t + 1}{3 \sqrt{x}+1}$$

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

$$\frac{dy}{dt}(\frac{ba +t}{a}) = -(\frac {y +4ab}{2b})$$

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

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

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

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

Last edited: Apr 23, 2006
2. Apr 23, 2006

### nrqed

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 !

Last edited: Apr 23, 2006
3. Apr 23, 2006

### GregA

Fixed the parenthisis
By setting t =0 in:$$x + 2x^{\frac {3}{2}} = t^2 + t$$
I get :$$x + 2x^{\frac {3}{2}} = 0$$

$$x^{\frac {1}{2}} = -\frac{1}{2}$$
x = 1/4

That response has just made me see how -6 is possible tho

Last edited: Apr 23, 2006
4. Apr 23, 2006

### nrqed

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. Apr 23, 2006

### GregA

Thankyou very much Nrged for putting my mind at ease and pointing out a solution I kept overlooking...what I did *lost* a solution!

Last edited: Apr 23, 2006
6. Apr 23, 2006

Actually, no. $x^{\frac{1}{2}}= -\frac{1}{2}$ has no solution since the 1/2 power of any real number is non-negative. The only root os $x+ 2x^{\frac{3}{2}}= 0 is 0. 7. Apr 23, 2006 ### GregA Thanks for pointing out where I am wrong HallsofIvy I won't make this same mistake twice. 8. Apr 23, 2006 ### nrqed So if we have x^2 = 4, the solution is only x=2 and cannot be x=-2? (where x is a real) ? 9. Apr 23, 2006 ### mathmike no because x^2 = 4 does give two roots. +2 and -2. -2*-2 = 4 2*2 = 4 10. Apr 23, 2006 ### nrqed I know, but HallsofIvy seems to imply the contrary which kind of surprised me . I'd like to understand what he meant (he is very knowledgeable so I must be missing his point entirely) 11. Apr 27, 2006 ### nrqed I am still confused by that statement: since the 1/2 power of any real number is non-negative. :surprised 12. Apr 27, 2006 ### HallsofIvy The equation x2= a has two solutions. They can be written [itex]\sqrt{a}$ and $-\sqrt{a}$. The reason why we need to write "+" and "-" is because $\sqrt{a}$ 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, $\sqrt{a}$, is defined as the positive solution to x2= a.
a1/2 is also [ b]defined[/b] as $\sqrt{a}$, the positive real root of x2[/sub]= a.

13. Apr 28, 2006

### nrqed

Ok, that makes sense. Thank you.