Question

A particle moves along the curve of y^3=2x+1 so that it's distance from the x-axis is increasing at a constant rate of 2 units/sec. When t=0, the particle is at (0,1).
(a) Find a pair of parametric functions x=x(t) and y=y(t) that describe the motion of the particle for nonnegative t.
(b) Find a , the the magnitude of the particle's acceleration, when t=1.

Answer 1

@jim_thompson5910

Answer 2

@iambatman

Answer 3

@Abhisar

Answer 4

@.Sam.

Answer 5

@TheSmartOne

Answer 6

@Destinymasha

Answer 7

Normally you let x = t, but because of that cube on the y, it'll probably be easier to make y = t.

So if y = t, then y^3 = t^3 and this is equal to 2x+1. Solve t^3 = 2x+1 for x to get

2x+1 = t^3

x = (t^3 - 1)/2

Answer 8

so

x = (t^3 - 1)/2
y = t

Answer 9

So that's all you have to do to find the parametric equations?

Answer 10

yeah I think so

Answer 11

there may be another way to do it though

Answer 12

How about the second part?

Answer 13

wait

Answer 14

I messed up

it says "When t=0, the particle is at (0,1). "

but if you plug t = 0 into y = t, you don't get y = 1

Answer 15

oh that's right

Answer 16

I'm not sure how to fix it, but I'll think of a way. One sec

Answer 17

ok fixed it

Answer 18

what you have to do is replace every t with t+1 to make sure we're at (0,1) when t = 0

Answer 19

so we have

x = ( (t+1)^3 - 1 )/2
y = t+1

Answer 20

ah i see

Answer 21

to find the acceleration, you need to compute x'' and y'' and plug in t = 1

Answer 22

so the parametric functions you just derived represented a position vector?

Answer 23

yeah

Answer 24

ok thanks a lot

Answer 25

np