Parametric equations - particle along a curve

[parlay]

This is a question on an AP Calc free-response problem I have, I have completed the first two steps and I'm stuck on the last two. The question is as follows:

A particle moves along a curve defined by the parametric equations x(t) = 2t and y(t) = 36-t^2 for time t, 0 ? t ? 6. A laser light on the particle points in the direction of motion and shines on the x- axis.

The first part of the question asks for the velocity vector of the particle, which I found to be <2, -2t>

The second part asks for the equation of the line tangent to the graph at (2t, 36-t^2) in terms of t and x, which I found to be
y= 3t^2 -tx +36

The other two parts are as follows:

Express the x-coordinate of the point on the x-axis that the light hits as a function of t.

At what time t is the light moving along the x-axis with the slowest speed? Justify your answer.

I really have no clue what to do for these last two parts, an explanation and the answer would be greatfully appreciated.

Sorry for the very long description. Answers very appreciated!

 

[fasteddie65]

This is a question on an AP Calc free-response problem I have, I have completed the first two steps and I'm stuck on the last two. The question is as follows:

A particle moves along a curve defined by the parametric equations x(t) = 2t and y(t) = 36-t^2 for time t, 0 ? t ? 6. A laser light on the particle points in the direction of motion and shines on the x- axis.

The first part of the question asks for the velocity vector of the particle, which I found to be <2, -2t>

The second part asks for the equation of the line tangent to the graph at (2t, 36-t^2) in terms of t and x, which I found to be
y= 3t^2 -tx +36

The other two parts are as follows:

Express the x-coordinate of the point on the x-axis that the light hits as a function of t.

At what time t is the light moving along the x-axis with the slowest speed? Justify your answer.

I really have no clue what to do for these last two parts, an explanation and the answer would be greatfully appreciated.

Sorry for the very long description. Answers very appreciated!

To find where the point is on the x-axis, let y = 0. y(t) = 36 - t^2 = 0 means that t = ±6, so consider t = 6. x(t) = 2t = 2(6) = 12.

v(t) = sqrt (x'(t)^2 + y'(t)^2) = sqrt (2^2 + (-2t)^2) = sqrt (4 + 4t^2)

To minimize v(t) is equivalent to minimizing v^2(t). To minimize v^2 = 4 + 4t^2 = V(t), V'(t) = 8t = 0. t = 0.
v(0) = sqrt 4 = 2, so the minimum speed is 2. The minimum speed occurs when t = 0.

 

[mmm4444bot]

… 0 ? t ? 6. A laser light [mounted] on the particle … shines on the x- axis …

Do you believe that this statement is true for all t?

I don't. I comment further, below, in my discussion of part(3).

… the equation of the line tangent to the graph … which I found to be

y = 3t^2 - tx + 36 …

For part (2), I got a different tangent-line equation, so I checked the case when t = 1 second.

The particle is at (2, 35) when t = 1.

The velocity vector tells us that the slope is -1 at those coordinates, Captain.

y - 35 = -(x - 2)

y = -x + 37

When I put t = 1 into your result, I get y = -x + 39.

When I put t = 1 into my result, I get y = -x + 37.

I'll say that I'm encouraged. (Denis might be lurking.)

Please check your work on part (2).

On part (3), I think that Fast Eddie might have misread the exercise because the location of the tangent line's x-intercept is not constant. In fact, when t = 0, I think that the tangent line never reaches the x-axis. Agree? Disagree?

Once you're sure that you have the correct equation for the tangent line in terms of x and t in part(2), then you can get an expression for the x-intercept by setting y equal to zero and solving for x.

As a confirmation, know that when t = 1, the x-intercept is 37, right? (When t = 1, the slope is -1. We see above that it crosses the y-axis at 37, so it must cross the x-axis at 37, too.)

To confirm, then, if we call the x-intercept function big X(t), then your result for part(3) must give the following.

X(1) = 37

As another confirmation, X(0) must be undefined, if you agree that there is no x-intercept when t = 0.

On part (4), I approached it with different reasoning than Eddie.

I considered the velocity of the particle in terms of the velocity vector <2, -2t>.

There's two dimensions to the movement. The horizontal component never changes, so it makes no contribution that would alter the particle's velocity.

Therefore, any change in velocity must arise from the vertical component, alone.

The vertical component is a linear function of time with positive slope. In other words, the higher the value of t, the higher the vertical "speed".

Clearly, the overall velocity is lowest when t = 0.