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I appreciate this is probably simpler than most of the questions on this site but here goes. I am doing a practice paper as a revision for a calculus exam coming up shortly. As we have already covered most of the following further calculus I thought this would be easy until I came to these two questions:

Question 8

Find the stationary point(s) for the following curve. For each that you find, determine whether it is a maximum or a minimum.

$y = \sqrt x(x − 8)$

Question 9

Find the equation of the normal to the curve at the point where $x = 1$

$y = \frac{(2\sqrt x+1)^2}{(3\sqrt x)}$

As I have typed this question I have realised that I could have used the product rule for 8 and the quotient rule for 9. BUT neither of these are a requirement of this exam so I am either missing something obvious or these questions shouldn't be on the paper. Either way I am still struggling.

I have got as far as:

Q8: $\frac{dy}{dx} = \frac{3}{2}x^{1/2} - 4x^{-1/2} = 0$ at the stat points

Q9: $\frac{dy}{dx} = \frac{2}{3}x^{-1/2} - \frac{1}{6}x^{-3/2}$

Can someone please show me the full working out?

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  • $\begingroup$ Welcome to MSE. Please use MathJax to format math on this site. To begin with, enclose all math expressions (including numbers) in $ signs. For example, $x_1^2$ will give you $x_1^2$. You'll get a much better response if your posts are easy to read. $\endgroup$
    – saulspatz
    Apr 12, 2021 at 19:28
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    $\begingroup$ You don’t have to use the product/quotient rules. For the first one, just expand the bracket and for the second, use the formula for $(a+b)^2$ and then divide each term by the denominator. $\endgroup$
    – Vishu
    Apr 12, 2021 at 19:31
  • $\begingroup$ "I could have used the product rule for 8 and the quotient rule for 9". Personally, assuming that I am interpreting your non-MathJax expressions correctly, I would invoke the product rule and the quotient rule, as a direct (inelegant) line of attack. $\endgroup$ Apr 12, 2021 at 19:32
  • $\begingroup$ Again sorry about the non MathJax, I've don't my best for now. I'll keep working on it. Never even heard of it until today. $\endgroup$ Apr 12, 2021 at 19:40
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    $\begingroup$ I edited your formulas. Please check that I haven't accidentally altered the meaning. $\endgroup$
    – saulspatz
    Apr 12, 2021 at 19:58

1 Answer 1

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Hint: now you know what is the coefficient of the normal line, since the product of the slopes of two perpendicular lines is -1, then you just need to plot $x = 1$, and take the inverse of the coeffiecient given by $dy/dx $ in Q9. You also know that the equation of line is $y - y_1 = m(x - x_1)$, but since the two lines have intersection point, then $x_1, y_1$ can be found from the equation originally given, for $x_1 = 1$, you get $y_1$.

For the stationary points, it's quite simple, you can replace $y^{-1}$ by $y$ and multiply by $y$, and we have:

$$\frac{3y^2}{2} = 4$$

then, from here you get 2 solutions.

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