Question

A particle moves on the curve y = 3e^(x^2 + 1). Find the rate of change of the y-coordinate at the instant the x-coordinate is 2 ad the x-coordinate is changing at 1/12 units/second.

Can someone please post the solution to this problem. I will give a medal.

Answer 1

looks like you are suppose to take x' as 1/12 instead of 1

Answer 2

So before we consider x' as something other than 1
do you know how to find dy/dx?

Answer 3

dy/dx would just be 6xe^(x^2+1), right?

Answer 4

\[\text{ so } y'=3e^{x^2+1}(2x)x' \\y'=6xe^{x^2+1}x' \\ y'=6xe^{x^2+1} \cdot \frac{1}{12} \\ y'=\frac{1}{2} xe^{x^2+1}\]
Find y' at x=2

Answer 5

On the first line you wrote that \[y' = 3e ^{x^2+1} (2x) x'\], how did you get x' in the equation?

Answer 6

x' is the rate of x

Answer 7

x' isn't 1 here

Answer 8

x' is 1/12

Answer 9

ok, apparently the derivative that I originally got did not have x' in it, can you please explain to me what I had done wrong?

Answer 10

\[D(x,y)=\sqrt{x^2+y^2} \\ D^2=x^2+y^2 \\ \text{ by chain rule we have } \\ 2D \frac{dD}{dt}=2x \frac{dx}{dt}+2y \frac{dy}{dt} \\ D \frac{dD}{dt}=x \frac{dx}{dt}+2 y \frac{dy}{dt}\]
and we just found dy/dt

Answer 11

dx/dt is given

Answer 12

oops that equation shouldn't have that 2 in it

Answer 13

so we are taking the derivative in respect to t?

Answer 14

\[D(x,y)=\sqrt{x^2+y^2} \\ D^2=x^2+y^2 \\ \text{ by chain rule we have } \\ 2D\frac{dD}{dt}=2x \frac{dx}{dt}+2y \frac{dy}{dt} \\ D \frac{dD}{dt}=x \frac{dx}{dt}+y \frac{dy}{dt}\]

Answer 15

yep

Answer 16

x'=dx/dt and y'=dy/dt

Answer 17

there has to be another variable in which x depends since x has rate other than 1

Answer 18

ahhhhh, that means that we have to do implicit differentiation in this problem, that makes sense. Thanks for the help!