# Tangent line to a parameterized curve implicitly defined?

Let

$$F=\sin(xe^y)+e^y\cos z-1$$

and

$$G=x^2-e^{xz}-z+1$$.

Consider the system

$$\left\{\begin{matrix} F(x,y,z)=0& & \\ G(x,y,z)=0& & \end{matrix}\right.$$

(i) Prove that the system implicitly defines a curve in space, in a neighbourhood of $$(0,0,0)$$. (ii) Calculate the tangent to the curve at the origin.

Applying two times the implicit function theorem, we end up writing the solutions to the system near $$(0,0,0)$$ as $$(x,α(x),β(x))$$, which is a curve with parameter $$x$$, and the first point (i) is done. We also have $$α'$$ and $$β'$$.

But how can we approach the second? I found that the direction of the tangent line in such a case should be parallel to the vector given by the cross product between the gradients of the two functions. Is this true? Why?

If we find the direction of the tangent, can we also find its equation in this problem?

• Actually you can consider the function $$\mathbf f:\mathbb R^3\to \mathbb R^2, \mathbf x\mapsto (F(\mathbf x),G(\mathbf x))$$ and then apply the Implicit function Theorem on $\mathbf f$. You can get an explicit formula for the derivatives of $\alpha$ and $\beta$ for example here: en.wikipedia.org/wiki/… – Maximilian Janisch Dec 28 '19 at 12:15
• Thanks @MaximilianJanisch. Actually I have the formulas for the derivatives and have done the first point. What I'm missing is how to use this information to find the line tangent to the curve of the solutions – Shootforthemoon Dec 28 '19 at 12:23
• The following might be helpful: If you have some interval $I\subset \mathbb R$ and a curve $$\gamma:I\to\mathbb R^n,$$ then you get a tangent vector to $\gamma$ at $x$ simply by $\dot\gamma(x)$. For example, in your case $$\gamma(x)=(x,\alpha(x),\beta(x))$$ so a tangent vector is $$\left.\frac{\mathrm d}{\mathrm dx}\right|_{x=0} (x,\alpha(x),\beta(x))=(1,\alpha'(0),\beta'(0))$$ – Maximilian Janisch Dec 28 '19 at 12:26
• Thanks very much for the help! I was struggling with this question ahaha – Shootforthemoon Dec 28 '19 at 12:38
• The tangent line must perforce lie on each of the tangent planes to the two implicitly-defined surfaces. – amd Dec 28 '19 at 18:50

For the second part, you may approximate $$F=G=0$$ near $$(0,0,0)$$ as follows,

$$F(x, y,z) \approx x+(1+y)-1 = x+y=0$$

$$G(x,y,z)\approx -z =0$$

Their normal vectors at origin are $$(1,1,0)$$ and $$(0,0,-1)$$ respectively. Thus, the tangent is $$(1,1,0)\times (0,0,-1)=(-1,1,0)$$ at the origin.

• Thanks! However, I quite don't get how the tangent is the product between the gradients – Shootforthemoon Dec 28 '19 at 12:20
• @Shootforthemoon - Visualize that the tangent of their intersection is perpendicular to the normal vectors of both surfaces. – Quanto Dec 28 '19 at 12:25
• Thanks, and do we have only the slope of the tangent or also the entire equation? – Shootforthemoon Dec 28 '19 at 12:27
• That is correct – Quanto Dec 28 '19 at 14:18
• @Shootforthemoon If you want to be pedantic one can note that we are in $\mathbb R^3$. So the equation $y=-x$ gives you a plane and the two equations $y=-x$ and $z=0$ together give you the line you want – Maximilian Janisch Dec 28 '19 at 15:03