Multivariable calculus (implicit function)

**kareman** · March 29th 2012, 01:51 AM

Let z a function of x,y near (2,1,-1) that is determined by $x^2z+xyz^3=-6$ . If $x(t)=t^3-3t$ and $y(t) =3t^2-6t+1$ , find $\frac{dz}{dt}$ for $t=2$ .

**Prove It** · March 29th 2012, 02:28 AM

Originally Posted by kareman

Let z a function of x,y near (2,1,-1) that is determined by x^2*z+x*y*z^3=-6. If x(t)=t^3-3t and y(t) =3t^2-6t+1, find dz/dt for t=2.

1. Substitute the functions of t into z.

2. Differentiate both sides implicitly with respect to t.

3. Solve for dz/dt.

**kareman** · March 29th 2012, 07:55 AM

Thank you for your response. Could you please check the following?

1. Substitute the functions of t into z.

$(t^3-3t)^2z+(t^3-3t)(3t^2-6t+1)z^3=-6$

2. Differentiate both sides implicitly with respect to t.

$2(t^3-3t)(3t^2-3)z + (t^3-3t)^2 \frac{dz}{dt} + \frac{d[(3t^5-6t^4+t^3-9t^3+18t^2-3t)z^3 ]}{dt}=0$

For t=2:
$2(2^3-3(2))(3(2)^2-3)z + (2^3-3(2))^2 \frac{dz}{dt} +\frac{d[(3(2)^5-6(2)^4+2^3-9(2)^3+18(2)^2-3(2)) z^3]}{dt}=0\Rightarrow2z(2)(9)+4\frac{dz}{dt}+\frac{d[(96-96+8-72+72-6) z^3 ]}{dt}=0\Rightarrow36z+4\frac{dz}{dt}+\frac{d[2z^3]}{dt}=0$

3. Solve for $\frac{dz}{dt}$ .
$36z + 4 \frac{dz}{dt} + 6z^2\frac{dz}{dt}=0\Rightarrow\frac{dz}{dt}=\frac{-36z}{6z^2+4}$

For z=-1:
$\frac{dz}{dt} (z=-1) = \frac{-36(-1)}{6(-1)^2+4}=\frac{36}{10}=3.6$

**Prove It** · April 1st 2012, 03:13 AM

Originally Posted by kareman

Thank you for your response. Could you please check the following?

1. Substitute the functions of t into z.

$(t^3-3t)^2z+(t^3-3t)(3t^2-6t+1)z^3=-6$

2. Differentiate both sides implicitly with respect to t.

$2(t^3-3t)(3t^2-3)z + (t^3-3t)^2 \frac{dz}{dt} + \frac{d[(3t^5-6t^4+t^3-9t^3+18t^2-3t)z^3 ]}{dt}=0$

For t=2:
$2(2^3-3(2))(3(2)^2-3)z + (2^3-3(2))^2 \frac{dz}{dt} +\frac{d[(3(2)^5-6(2)^4+2^3-9(2)^3+18(2)^2-3(2)) z^3]}{dt}=0\Rightarrow2z(2)(9)+4\frac{dz}{dt}+\frac{d[(96-96+8-72+72-6) z^3 ]}{dt}=0\Rightarrow36z+4\frac{dz}{dt}+\frac{d[2z^3]}{dt}=0$

3. Solve for $\frac{dz}{dt}$ .
$36z + 4 \frac{dz}{dt} + 6z^2\frac{dz}{dt}=0\Rightarrow\frac{dz}{dt}=\frac{-36z}{6z^2+4}$

For z=-1:
$\frac{dz}{dt} (z=-1) = \frac{-36(-1)}{6(-1)^2+4}=\frac{36}{10}=3.6$

You should evaluate the derivative fully before substituting any points...

**biffboy** · April 1st 2012, 04:28 AM

Might be quicker to differentiate both sides of given equation with respect to x. Hence find dz/dx then use dz/dt=dz/dx times dx/dt
dy/dx will occur but you can get that from dy/dt and dx/dt

**kareman** · April 4th 2012, 11:37 PM

Originally Posted by Prove It

You should evaluate the derivative fully before substituting any points...

Yes, I should have evaluated the derivative first! The right solution is the following:

$2(t^3-3t)(3t^2-3)z+(t^3-3t)^2\frac{dz}{dt}+(15t^4-24t^3+3t^2-27t^2+36t-3)z^3+(3t^5-6t^4+t^3-9t^3+18t^2-3t)3z^2\frac{dz}{dt}=0$

Now substituting t=2 gives:

$2(2^3-3(2))(3(2)^2-3)z+(2^3-3(2))^2\frac{dz}{dt}+(15(2)^4-24(2)^3+3(2)^2-27(2)^2+36(2)-3)z^3+(3(2)^5-6(2)^4+(2)^3-9(2)^3+18(2)^2-3(2))3(-1)^2\frac{dz}{dt}$

$=>36z+4\frac{dz}{dt}+21z^3+6z^2\frac{dz}{dt}$

Solving for $\frac{dz}{dt}$ :

$\frac{dz}{dt}=\frac{-21z^3-36z}{6z^2+4}$

and for z=-2:

$\frac{dz}{dt}=\frac{-21(-1)^3-36(-1)}{6(-1)^2+4}=\frac{57}{10}=5.7$

**kareman** · April 4th 2012, 11:47 PM

Originally Posted by biffboy

Might be quicker to differentiate both sides of given equation with respect to x. Hence find dz/dx then use dz/dt=dz/dx times dx/dt
dy/dx will occur but you can get that from dy/dt and dx/dt

I found it easier to set the function $F(x,y,z)=x^2z+xyz^3+6=0$ and use the chain rule:

$\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}=(2xz+yz^3)(3t^2-3)+(xz^3)(6t-6)+(x^2+3xyz^2)\frac{dz}{dt}=0$

Solving for $\frac{dz}{dt}$ gives

$\frac{dz}{dt}=-\frac{(2xz+yz^3)(3t^2-3)+(xz^3)(6t-6)}{x^2+3xyz^2}$

At point (2,1,-1) and for t=2 the result is

$\frac{dz}{dt}=5.7$

which is the same as the above method.

**biffboy** · April 5th 2012, 12:04 AM

Yes indeed. And you could have shortened this a little. You didn't need to write out the expression for dz/dt as the numbers could have gone straight in to the previous line.

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