# calculus BC slope

posted by
**Anonymous** on
.

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 ?