calculus BC slope
posted by Anonymous .
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.
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 ?