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If I have to work out the gradient of the tangent given the equation of a curve, after differentiating the equation of the curve, is it ok to use any x value on the tangent to work out its gradient??

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For example in this question, the tangent passes through (0,0), so can I substitute 0 into the differentiated curve equation to get the gradient of the tangent or does it have to be the x value of P?

Newbie101
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  • Can you give the entire problem, please? I'm not certain I understand what you are asking? – saulspatz Sep 30 '19 at 20:31
  • Differentiation gives the tangent's gradient immediately. You just need to specify the point at which you have a tangent. I'm not clear what you're trying to do. – J.G. Sep 30 '19 at 20:32

2 Answers2

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Suppose $P$ has $x=a$ so $OP$ has gradient $\frac32(3a-5)^{-1/2}$. Since the tangent passes through $O$, $(3a-5)^{1/2}=\frac32a(3a-5)^{-1/2}$, i.e. $3a-5=\frac32a$. You can solve for $a$ now.

J.G.
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  • Thanks for the answer, could you explain why passing through (0,0) means $(3−5)^1/2=3/2(3−5)^−1/2 $ – Newbie101 Sep 30 '19 at 20:49
  • @Newbie101 The left-hand side is $y$, the right-hand side $xy^\prime$. I'm using the same strategy you had in mind. – J.G. Sep 30 '19 at 20:58
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There exists a line through the origin that is tanget to the curve.

$y = mx$ is an equation through the origin.

$y = \sqrt {3x-5}$

We can do this without calculus. Find the places that the two curves intersect and find m such that there is one solution.

$y^2 = 3x - 5$

Substitute $y = mx$ on the LHS

$m^2x^2 -3x + 5 = 0$

use the quadratic formula:

$x = \frac 3 \pm \sqrt {9 - 20m^2}{2m^2}$

If there is one solution

$\sqrt {9 - 20m^2} = 0\\ m = \sqrt {\frac {9}{20}} = \frac {3}{2\sqrt 5} = \frac {3\sqrt 5}{10}\\ y = \frac{3\sqrt 5}{10} x$

If you want to use calculus

Find the derivative:

$\frac {dy}{dx} = \frac {3}{2\sqrt{3x - 5}}$

Now we need to find a line such that the slope of the line equals $\frac {dy}{dx}$ and intersects our curve at the appropriate point.

$\frac {y}{x} = \frac {\sqrt{3x - 5}}{x} = \frac {3}{2\sqrt{3x - 5}}\\ 3x - 5 = \frac 32 x\\ x = (\frac 23)(5) \frac {dy}{dx} = \frac {3}{2\sqrt5} $

And we have the same slope as above.

Which way did you think was easier?

Doug M
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