Question

how do you find the slope of the line tangent to the curve at the point where x=3 using the curve x=e^t and y=te^-t

Answer 1

Find the derivative of <e^t, te^-t>

Answer 2

the answer is supposed to be -.011 i tried dy/dt over dx/dt and plugged in 3 but kept getting it wrong

Answer 3

The line is the original vector evaluated at the point + the derivative vector evaluated at the point.

Answer 4

so what would that look like?

Answer 5

nevermind figured it out thanks for the help