How to Find the Tangent Vector to a Path under Vec Calc Transformation?

[tandoorichicken]

The Problem: No clue how to begin the following. The textbook isn't too clear.

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 ; (x,y) \rightarrow (e^{x+y}, e^{x-y})$ Let c(t) be a path with c(0)=(0,0) and c'(0)=(1,1). What is the tangent vector to the image of c(t) under f at t=0?

Any help at all would be much appreciated.

 

[benorin]

Let $$\vec{c}(t) = \left< x(t), y(t)\right>$$

then the image of c(t) under f is

$$f( \vec{c}(t) ) = f(x(t), y(t)) = \left( e^{x(t)+y(t)},e^{x(t)-y(t)}\right)$$

The tangent vector is

$$\frac{d}{dt} f( \vec{c}(t) ) =\frac{d}{dt} f(x(t), y(t)) = \frac{d}{dt}\left( e^{x(t)+y(t)},e^{x(t)-y(t)}\right) = \left( \frac{d}{dt} e^{x(t)+y(t)}, \frac{d}{dt} e^{x(t)-y(t)}\right)$$
$$= \left( e^{x(t)+y(t)}\left( x^{\prime}(t)+y^{\prime}(t)\right) , e^{x(t)-y(t)}\left( x^{\prime}(t)-y^{\prime}(t)\right) \right)$$

evalute at t=0 to get

$$\left( e^{x(0)+y(0)}\left( x^{\prime}(0)+y^{\prime}(0)\right) , e^{x(0)-y(0)}\left( x^{\prime}(0)-y^{\prime}(0)\right) \right)$$

recall that c(0)=(0,0) and c'(0)=(1,1) and hence x(0)=0, y(0)=0, x'(0)=1, and y'(0)=1 so that

$$\left( e^{x(0)+y(0)}\left( x^{\prime}(0)+y^{\prime}(0)\right) , e^{x(0)-y(0)}\left( x^{\prime}(0)-y^{\prime}(0)\right) \right) = \left( e^{0+0}\left( 1+1\right) , e^{0-0}\left( 1-1\right) \right) = \left( 2 , 0\right)$$

is the tangent vector to the image of c(t) under f at t=0.

Though I'm not sure... is f a vector valued function of vectors?

 

[HallsofIvy]

Though I'm not sure... is f a vector valued function of vectors?

Yes, that's exactly what $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 ; (x,y) \rightarrow (e^{x+y}, e^{x-y})$ means.

And your solution is completely correct. It's really an exercise in using the chain rule.

 

[benorin]

Thanx Ivy.