I'm stuck on the following question and don't know how the book got its answer because it has multiple variables (its precalc so I don't know how to do calculus yet) and I don't know how to eliminate one completely. Any help is much appreciated.

Question: Suppose Fritzy the Fox, positioned at a point $(x,y)$ in the first quadrant, spots Chewbacca the Bunny at $(0,0)$. Chewbacca begins to run along a fence (the positive $y$-axis) towards his warren. Fritzy, of course, takes chase and constantly adjusts his direction so that he is always running directly at Chewbacca. If Chewbacca's speed is $v_1$ and Fritzy's speed is $v_2$, the path Fritzy will take to intercept Chewbacca, provided $v_2$ is directly proportional to, but not equal to $v_1$ is modelled by: $$ y = \frac{1}{2}\left(\frac{x^{1+v_1/v_2}}{1+v_1/v_2}-\frac{x^{1-v_1/v_2}}{1-v_1/v_2}\right) + \frac{v_1v_2}{v_2^2-v_1^2} $$

a.) Determine the path that Fritzy will take if he runs exactly twice as fast as Chewbacca; that is $v_2$ = $2v_1$. Use your calculator to graph this path for $x \ge 0$. What is the significance of the y-intercept of the graph?

So in order to graph this with the provisions of a (where $v_2$ = $2v_1$) I believe I need to reduce this function down to one variable. So here is my attempt: $$ y = \frac{1}{2}\left(\frac{x^{3v_1/2v_1}}{3v_1/2v_1}-\frac{x^{v_1/2v_1}}{v_1/2v_1}\right) + \frac{3v_1}{v_1^2} $$ $$ y = \frac{1}{2}\left(\frac{x^{3v_1/2v_1} - 3x^{v_1/2v_1}}{3v_1/2v_1}\right) + \frac{3v_1}{v_1^2} $$ From here I believe I can get rid of the $v_1$ in the first term because it is present in both numerators and denominators. So here goes: $$ y = \frac{1}{2}\left(\frac{x^{3/2} - 3x^{1/2}}{3/2}\right) + \frac{3v_1}{v_1^2}. $$ But I don't know how you get rid of the $v_1$ in the second term because it is squared in the denominator.

I maybe even be on the completely wrong path as the book answer is: $$ y = \frac{1}{3} x^{3/2} - \sqrt{x} + \frac{2}{3}. $$

Can anyone help me understand what I have to do to get the book answer?

  • 4
    $\begingroup$ Please use a more specific title $\endgroup$ – Viktor Glombik Apr 22 '19 at 14:08

You are correct apart from the last fraction $$v_1v_2=v_1(2v_1)=2v_1^2$$ $$v_2^2-v_1^2=(2v_1)^2-v_1^2=4v_1^2-v_1^2=3v_1^2$$ So $$\frac{v_1v_2}{v_2^2-v_1^2}=\frac{2v_1^2}{3v_1^2}=\frac23$$ also your answer simplifies becuase $$\frac12\left(\frac{x^\frac32-3x^\frac12}{3/2}\right)=\frac{x^\frac32-3x^\frac12}{3}=\frac13x^\frac32-x^\frac12=\frac13x^\frac32-\sqrt{x}$$

  • $\begingroup$ Thank you Peter! $\endgroup$ – maybedave Apr 22 '19 at 14:19

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