1

Curve and a tangent y=25/4 is given along with the question here

The equation of the curve is $$y=x\left(\frac{k}{\sqrt{x}} - 1\right)$$

Does the problem mean the curve has a slope of zero at $y = 25/4$? The problem asks to find the value of $k$ and equation of line "l" which can be seen in the graph.

PM 2Ring
  • 4,844
netdev2000
  • 15
  • Does the problem mean the curve has a slope of zero at y = 25/4? Yes. And that the curve just touches that line, it doesn't cross it. – PM 2Ring Aug 26 '21 at 05:45
  • dy/dx = (k/(2√x)) - 1. How to take it forward from there? Can you please solve the problem, find k and equation of line "l"? I am getting something like k√x-x-25/4= k/(2√x)) - 1. But I am not getting k from that. Answer is k=5. – netdev2000 Aug 26 '21 at 10:34
  • Yes, that derivative is correct. But you should put that into your question. And add that 2nd equation too, and explain how / why you got it. – PM 2Ring Aug 26 '21 at 11:18

1 Answers1

0

Note that the equation of the curve

$$y = x\left(\frac{k}{\sqrt x} - 1\right) \tag 1$$

is undefined at $x=0$. We can change it to the more convenient

$$y = k\sqrt x - x \tag 2$$

which is equivalent, except that it is defined at zero. As you mention in the comments,

$$y' = \frac{dy}{dx} = \frac{k}{2\sqrt x} - 1 \tag 3$$

We're given that the curve is tangent to the horizontal line

$$y = 25/4 \tag 4$$

We can find where the curve has a horizontal tangent by setting $y'=0$.

$$\frac{k}{2\sqrt x} = 1$$ $$\sqrt x = \frac{k}{2}$$ $$x = \frac{k^2}{4}$$

Now we can plug that $\sqrt x$, $x$, and $y=25/4$ into eqn (2).

$$25/4 = k \frac{k}{2} - \frac{k^2}{4}$$ $$25/4 = \frac{k^2}{2} - \frac{k^2}{4}$$ $$25/4 = \frac{k^2}{4}$$ Thus $k=5$

For part (ii) of the question, we can see from eqn (3) that for large $x$, the slope of the tangent line to the curve approaches -1 because $\frac{k}{2\sqrt x}$ is small. Eg, at $x=10000$, $y'=\frac{1}{200} - 1$.

However, the tangent line is displaced from the line $y = -x$. In fact, at $x = x_0$, the equation of the tangent line is

$$y = \left(\frac{5}{2\sqrt {x_0}}-1\right)x + \frac52\sqrt{x_0}$$

PM 2Ring
  • 4,844
  • Sorry for late reply. Thank you. I missed that I can get x in terms of k from (k/2√x) - 1 = 0 which can then be plugged back to equation of the curve y = k√x - x when y = 25/4. Also the part (i i) is nicely explained. As for the last line of your answer ( y = mx + (k/2√x)) , the equation of the tangent line, I am trying to figure out how it equates to y-y1 = m (x-x1). I do see the curve passes through (0,0) though. Hence I am getting y-0 = m(x-0) , i.e. y = ((5/2√x)-1).x. Thanks again. – netdev2000 Sep 12 '21 at 09:44
  • @netdev2000 That final tangent equation can be written as: $$y-\left(5\sqrt{x_0}-x_0\right) = \left(\frac{5}{2\sqrt {x_0}}-1\right)\left(x-x_0\right)$$ – PM 2Ring Sep 12 '21 at 13:02
  • The curve in equation (1) does not pass through $(0, 0)$. It's undefined for $x\le 0$. – PM 2Ring Sep 12 '21 at 13:04
  • Ok, got it, and thanks again.. – netdev2000 Sep 13 '21 at 05:42