Question

Help with 4b thank you. ​

Answer 1

First let's compute dx/dt

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

Now compute dy/dt

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

From here, apply the chain rule to say

$\frac{dy}{dx} = \frac{dy*dt}{dx*dt}\\\\\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\\\\\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\\\\\frac{dy}{dx} = \frac{2t^2-1}{t^2} \div \frac{t^2+1}{t^{2}}\\\\\frac{dy}{dx} = \frac{2t^2-1}{t^2} \times \frac{t^{2}}{t^2+1}\\\\\frac{dy}{dx} = \frac{2t^2-1}{t^2+1}$

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

$\frac{dy}{dx} = \frac{2t^2-1}{t^2+1}\\\\\frac{dy}{dx} = \frac{2t^2-1+2-2}{t^2+1}\\\\\frac{dy}{dx} = \frac{(2t^2+2)-1-2}{t^2+1}\\\\\frac{dy}{dx} = \frac{2(t^2+1)-3}{t^2+1}\\\\\frac{dy}{dx} = \frac{2(t^2+1)}{t^2+1}-\frac{3}{t^2+1}\\\\\frac{dy}{dx} = 2-\frac{3}{t^2+1}$

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,

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

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 $-1 < \frac{dy}{dx}$

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

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

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

If t approaches positive infinity, then,

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

As t approaches infinity, the dy/dx value approaches L = 2 from below.

The same applies when t approaches negative infinity.

So we see that $\frac{dy}{dx} < 2$

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Since $-1 < \frac{dy}{dx} \text{ and } \frac{dy}{dx} < 2$, those two inequalities combine into the compound inequality $-1 < \frac{dy}{dx} < 2$

So dy/dx is bounded between -1 and 2, exclusive of either endpoint.