Calculus on Vectors: Parametric Equations and Initial-Value Problems"

[johniie77]

Homework Statement

1)Find parametric equations of the line tangent to the graph of r(t) at the point where t=t0
r(t)=ln(t) i + e^-t j +t^3 k; t0=2

2) solve the vector initial-value problem for y(t) by integrating and using the initial conditions to find the constants of integration.
y''(t) = 12T^2 i - 2t j, y(0)=2i-4j y'(0)=0

The Attempt at a Solution

I just wanted to make sure i am doing these right
1) x=ln(2)+1 y=0 z=20
2) (T^4 +2)i +(-(T^3/3)-4)j

 

[LCKurtz]

Homework Statement

1)Find parametric equations of the line tangent to the graph of r(t) at the point where t=t0
r(t)=ln(t) i + e^-t j +t^3 k; t0=2

2) solve the vector initial-value problem for y(t) by integrating and using the initial conditions to find the constants of integration.
y''(t) = 12T^2 i - 2t j, y(0)=2i-4j y'(0)=0

The Attempt at a Solution

I just wanted to make sure i am doing these right
1) x=ln(2)+1 y=0 z=20
2) (T^4 +2)i +(-(T^3/3)-4)j

The first answer describes a point, not the equation of a line.
The second can be checked by plugging 0 into the answer and its derivatives and checking it.

 

[flyingpig]

1) Do you know the equation for tangent here?

 

[johniie77]

1) yea its just the derivative,
@Lckurtz do you mean this? x=ln(T) + 1 y=e^-T - e^-T z=T^3 +2*T^2

2) not sure what you mean, if i plug in zero into the answer then i just get 2i -4j?

 

[flyingpig]

1) yea its just the derivative,
@Lckurtz do you mean this? x=ln(T) + 1 y=e^-T - e^-T z=T^3 +2*T^2

No it's a line, you have a lot of nonlinear terms

 

[SammyS]

1) yea its just the derivative,
@Lckurtz do you mean this? x=ln(T) + 1 y=e^-T - e^-T z=T^3 +2*T^2

LCKurtz is not online right now so I'll answer.

No. As the flyingpig said, does NOT describe a line. However, it is parametric.

2) not sure what you mean, if i plug in zero into the answer then i just get 2i -4j?

Yes, you do get 2i -4j for y(0).

What do you get for y'(0) >

 

[johniie77]

LCKurtz is not online right now so I'll answer.

No. As the flyingpig said, does NOT describe a line. However, it is parametric.

huh that's all i thought the question asked for, for the parametric equations? So what is the question looking for?

Yes, you do get 2i -4j for y(0).

What do you get for y'(0) >

you get 0 for y'.

 

[flyingpig]

What you have don't even look like a line

Look what you have

y=e^-T - e^-T

That's y = 0!

 

[johniie77]

oh, i think i was simplying it too much, so do you mean this?
r=(ln(T) +1) i +(e^-T-e^-T) j +(T^3 +3T^2)k

 

[flyingpig]

oh, i think i was simplying it too much, so do you mean this?
r=(ln(T) +1) i +(e^-T-e^-T) j +(T^3 +3T^2)k

No, don't even use r(t) again.

T^3 + 3T^2 what is this? This is a curve, you know, that thing which isn't straight?

 

[SammyS]

What does r'(t) give you when t = 2 ?

 

[johniie77]

oh i think i get what you guys are saying, i plugged into r(2) and solved for that and then r'(2) and got
r(t) = <ln(2),e^-2,8> +T<.5,e^-2,12>, which is an equation of a line and is parametric.

 

[SammyS]

Yes, that's more like it !

 

[johniie77]

cool finally understand what I was doing wrong, and as for number 2 you said to plug in 0 into the equation and its derivatives which i did, but I am not sure how that tells me if i did it right
2i -4j for y(0).
0 for y'.
and 0 for y''
?

 

[SammyS]

One of the conditions to be satisfied for y is that y'(0) = 0. So from that point of view, you solution checks out.