Question

An object moving along a curve in the xy-plane has position (x(t),y(t)) at time t>=0 with dx/dt=3+cos(t^2). The derivative dy/dt is not explicitly given. At time t=2, the object is at position (1,8).

(a) Find the x-coordinate of the position of object at time t=4.

(b) At time t=2, the value of dy/dt is -7. Write an equation for the line tangent to the curve at the point (x(2),y(2))

(c) Find the speed of the object at time t=2.

(d) For t>=3, the line tangent to the curve at (x(t),y(t)) has a slope of 2t+1. Find acceleration vector of the object at time t=4.

Answer 1

Please give me hints rather than the entire solution.

Answer 2

a) integrate 3+cos(t^2) dt ... use initial value of x(2) = 1

Answer 3

I have to integrate it?

Answer 4

yes .. integrate if form 2 to 4

Answer 5

I don't know how to integrate cos(t^2)

Answer 6

are you allowed to use calculator?? i mean some sorta graphic calculator or .. wolfram??
this is definite integral ... my guess is ... indefinite integral is not closed or non elementary
http://www.wolframalpha.com/input/?i=integrate+cos%28x^2%29

Answer 7

add 1 to this http://www.wolframalpha.com/input/?i=integrate+cos%28x^2%29%2B3+from+2+to+4

Answer 8

b) ... calculate the dx/dt at t=2 .. ie. dx/dt=3+cos(t^2)|t=2

dy/dx = dy/dt* 1/(dx/dt) <=== this slope m,
since you know coordinate 1,8
find tangent using single point formula

Answer 9

dy/dt is given, dx/dt ... you calculate in step b)

ds/dt = \( \sqrt{(dx/dt)^2 + (dy/dt)^2}\)

Answer 10

d) the acceleration is given by
d^2s/dt^2

dy/dx = 2t+1

dy/dt = (2t+1) dx/dt

use this formula to calculate ds/dt

differentate it again ...wrt t and put t=4

this will give your acceleration

Answer 11

Okay now I know why he only gave two questions for homework....
It is super difficult!!

Thank you, @experimentX !

Answer 12

well ... :)
good night!!