How to find the tangent line of a level curve of a function?

Question

I'm trying to solve this exercise but I'm having some issue.

Find the equation of the tangent line of $e^{x-y}(2x^2+y^2)$ at the point $(1,0)$ at the level curve.

So I start finding the gradient of the function
$gradf={e^{x-y}(2x^2+y^2)+4xe^{x-y} \choose -e^{x-y}(2x^2+y^2)+2ye^{x-y}}$

But now I don't know how to go on and even if find the gradient is actually correct, can someone help me by explaining me how to solve this excercise?

Answer 1

Let $$z = f(x,y) = e^{x-y} (2x^2 + y^2).$$ Then at the point $(1,0)$, the value of $z$ is $f(1,0) = 2e$, hence the level curve is given by the implicit relation $$f(x,y) = 2e = e^{x-y}(2x^2 + y^2).$$ Then implicit differentiation gives $$0 = \left(1 - \frac{dy}{dx}\right)e^{x-y}(2x^2 + y^2) + e^{x-y} \left(4x + 2y \frac{dy}{dx}\right),$$ or $$e^{x-y} \left(2x^2 + y^2 + 4x + (-2x^2 -y^2 + 2y)\frac{dy}{dx}\right) = 0$$ or $$\frac{dy}{dx} = \frac{2x^2 + 4x + y^2}{2x^2 + y^2 - 2y}.$$ At $(x,y) = (1,0)$, we find the slope $$\frac{dy}{dx} = \frac{6}{2} = 3$$ and the rest is trivial.

This solution is provided in order to illustrate that the concepts introduced in multivariable calculus are a natural extension of concepts from the univariate calculus; here, we treat the level set as an implicit curve--note it was not actually necessary to compute the $z$-value of that level set--and the implicit derivative computation is analogous to the gradient computation. This relationship is not a coincidence.

Answer 2

HINT

Recall that

• $\nabla f(x_0,y_0)$ is perpendicular to the level curve at $(x_0,y_0)$
• then the tangent vector $v$ at $(x_0,y_0)$ is such that $$v\cdot \nabla f(x_0,y_0)=0$$
• the parametric equation for the tangent line is $(x,y)=(x_0,y_0)+tv$
• let $v=(a,b)$ with $b\neq 0$ then the cartesian equation for the tangent is

$$(y-y_0)=\frac{b}{a}(x-x_0)$$

Answer 3

Your gradient looks correct to me.

Use the chain rule. Along the level curve $f(x,y) = c$, as long as $\frac{\partial f}{\partial y} \neq 0$, we can consider $y$ as implicitly a function of $x$. Then $$ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx} = 0 $$ so $$ \frac{dy}{dx} = - \frac{\partial f/\partial x}{\partial f/\partial y} $$