1 Answer

Geovany · Answer 1 · 2022-01-24T17:17:25+0000

First let's compute dx/dt

$x = t - (1)/(t)\\\\x = t - t^(-1)\\\\(dx)/(dt) = (d)/(dt)\left(t - t^(-1)\right)\\\\(dx)/(dt) = 1-(-1)t^(-2)\\\\(dx)/(dt) = 1+(1)/(t^(2))\\\\(dx)/(dt) = (t^2)/(t^(2))+(1)/(t^(2))\\\\(dx)/(dt) = (t^2+1)/(t^(2))\\\\$

Now compute dy/dt

$y = 2t + (1)/(t)\\\\y = 2t + t^(-1)\\\\(dy)/(dt) = (d)/(dt)\left(2t + t^(-1)\right)\\\\(dy)/(dt) = 2 - t^(-2)\\\\(dy)/(dt) = 2 - (1)/(t^2)\\\\(dy)/(dt) = (2t^2)/(t^2)-(1)/(t^2)\\\\(dy)/(dt) = (2t^2-1)/(t^2)\\\\$

From here, apply the chain rule to say

$(dy)/(dx) = (dy*dt)/(dx*dt)\\\\(dy)/(dx) = (dy)/(dt) * (dt)/(dx)\\\\(dy)/(dx) = (dy)/(dt) / (dx)/(dt)\\\\(dy)/(dx) = (2t^2-1)/(t^2) / (t^2+1)/(t^(2))\\\\(dy)/(dx) = (2t^2-1)/(t^2) * (t^(2))/(t^2+1)\\\\(dy)/(dx) = (2t^2-1)/(t^2+1)\\\\$

We could use polynomial long division, or we could add 2 and subtract 2 from the numerator and do a bit of algebra like so

$(dy)/(dx) = (2t^2-1)/(t^2+1)\\\\(dy)/(dx) = (2t^2-1+2-2)/(t^2+1)\\\\(dy)/(dx) = ((2t^2+2)-1-2)/(t^2+1)\\\\(dy)/(dx) = (2(t^2+1)-3)/(t^2+1)\\\\(dy)/(dx) = (2(t^2+1))/(t^2+1)-(3)/(t^2+1)\\\\(dy)/(dx) = 2-(3)/(t^2+1)\\\\$

This concludes the first part of 4b

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Now onto the second part.

Since t is nonzero, this means either t > 0 or t < 0.

If t > 0, then,

$t > 0\\\\t^2 > 0\\\\t^2+1 > 1\\\\(1)/(t^2+1) < 1 \ \text{ ... inequality sign flip}\\\\(3)/(t^2+1) < 3\\\\-(3)/(t^2+1) > -3 \ \text{ ... inequality sign flip}\\\\-(3)/(t^2+1)+2 > -3 + 2\\\\2-(3)/(t^2+1) > -1\\\\-1 < 2-(3)/(t^2+1)\\\\-1 < (dy)/(dx)\\\\$

note the inequality signs flipping when we apply the reciprocal to both sides, and when we multiply both sides by a negative value.

You should find that the same conclusion happens when we consider t < 0. Why? Because t < 0 becomes t^2 > 0 after we square both sides. The steps are the same as shown above.

So both t > 0 and t < 0 lead to
-1 < (dy)/(dx)

We can say that -1 is the lower bound of dy/dx. It never reaches -1 itself because t = 0 is not allowed.

We could say that

$\displaystyle \lim_(t\to0)\left(2-(3)/(t^2+1)\right)=-1\\\\$

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To establish the upper bound, we consider what happens when t approaches either infinity.

If t approaches positive infinity, then,

$\displaystyle L = \lim_(t\to\infty)\left(2-(3)/(t^2+1)\right)\\\\\\\displaystyle L = \lim_(t\to\infty)\left((2t^2-1)/(t^2+1)\right)\\\\\\\displaystyle L = \lim_(t\to\infty)\left((2-(1)/(t^2))/(1+(1)/(t^2))\right)\\\\\\\displaystyle L = (2-0)/(1+0)\\\\\\\displaystyle L = 2\\\\$