consider the curve in xy-plane by x=e^t and y=te^(-t) for t is greater or equal to o. the slope of the line tangent to the curve at the point where x=3.

all the answer choices are decimal points. a is 20.086, b is .342, c is -.005, d is -.011 and e is -.033

so what i did is that since these are parametric equations i used the formula dy/dx=(dy/dt)/(dx/dt) to get the derivative to get the slope. but when i use the y-y=m(x-X) equation to get line, how do i get y.

you are on the right track, I would have done the same thing.
lets check our derivatives.
I had
dx/dt = e^t and dy/dt = (e^-t)(1-t) after using the product rule for that one and then simplifying.

so my(dy/dt)/(dx/dt) = (1-t)/(e^2t)

I need the t value when x=3

3=e^t, so t=ln3

then dy/dx = (1-ln3)/e^(2ln3)
= -.0109.. which was one of the choices.

BTW, did you realize that e^(2ln3) = 9 ?

Wow, e^(2ln3) = 9? That's like multiplying by yourself and getting a totally different number. Talk about getting lost in the math! But hey, at least you found the slope you were looking for. And it looks like it's not the most positive slope in the world. In fact, it's a little negative, just like my bank account after splurging on clown shoes. So, the answer would be d) -.011. Now you can go back to the xy-plane knowing you've got the right slope to keep things on track.

Yes, you are correct. The value of e^(2ln3) is indeed 9. This is because the natural logarithm, ln, and the exponential function, e^x, are inverse functions of each other. So when we raise e to the power of 2ln3, it effectively cancels out the logarithm and gives us 3^2, which is 9.

Therefore, the correct answer for the slope of the tangent line at the point where x=3 is indeed -0.0109, which is closest to option d.