Why do these two seemingly equivalent equations not yield the same first derivative when differentiating with respect to $h$? This is the equation for the lateral surface area of a cone of radius $r$ and height $h$. I have been unable to yield equivalent derivatives. Thank you.

$4500\pi = \pi r(r^2 + h^2)^{1/2}$

$\frac{4500}r = (r^2 + h^2)^{1/2}$

edit: i actually resolved the issue myself


These expressions are not seemingly equivalent, they are exactly equivalent:

$$4500\pi = \pi r(r^2 + h^2)^{1/2}\implies\frac{4500\pi}{\pi r} = \frac{\pi r(r^2 + h^2)^{1/2}}{\pi r}\implies\frac{4500}r = (r^2 + h^2)^{1/2}$$

Likewise, the implicit derivative is the same:

$$4500 = r(r^2 + h^2)^{1/2}\implies\frac{dr}{dh}=-\frac{hr}{h^2+2r^2}$$

You can solve as follows:

$$4500^2=r^2(r^2+h^2)\implies (4500^2)'=(r^4)'+(r^2h^2)'\implies0=4r^3r'+(r^2\cdot h^2)'$$

$$0=4r^3r'+(r^2\cdot h^2)'\implies 0=4r^3r'+(r^2)'h^2+r^2(h^2)'=4r^3r'+2rr'h^2+2r^2h$$


$$-2r^2h=r'(4r^3+2rh^2)\implies\frac{-2r^2h}{4r^3+2rh^2}=r'\implies r'=\frac{2r(-hr)}{2r(2r^2+h^2)}=-\frac{hr}{h^2+2r^2}$$

's Just algebra and differentiation rules, man.


Solving without simplifying first gets you the same answer, it just more steps:

Let $k=4500$ (because I'm tired of writing the number).

$$\frac{k^2}{r^2}=r^2+h^2\implies k^2\left(\frac{1}{r^2}\right)'=2rr'+2h\implies-\frac{2 k^2 r'}{r^3}=2rr'+2h$$

$$-\frac{2 k^2 r'}{r^3}=2rr'+2h\implies-2k^2r'=2r^4r'+2hr^3\implies -2hr^3=r'(2r^4+2k^2)$$

$$-2hr^3=r'(2r^4+2k^2)\implies r'=-\frac{2hr^3}{2r^4+2k^2}$$

Substitute $k^2=r^4+h^2r^2$


  • $\begingroup$ I understand this, but when computing the derivative of the second equation ( 4500/r ), a different derivative is yielded, which I confirmed with an online derivative calculator. I would post the answer here but I am unfamiliar with how to format this properly. $\endgroup$
    – Ryan
    Oct 4 '19 at 1:18
  • $\begingroup$ @Ryan Sounds like an error. Wolfram Alpha gives the same implicit derivative for both equations. Because there are so many algebraic manipulations involved, you might encounter errors with the sign of the solution, complex/imaginary solutions, or division by zero - the same goes for calculators depending on how they're programmed. $\endgroup$
    – R. Burton
    Oct 4 '19 at 1:24
  • $\begingroup$ W.A.: wolframalpha.com/input/… and wolframalpha.com/input/… $\endgroup$
    – R. Burton
    Oct 4 '19 at 1:26
  • $\begingroup$ My professor corroborated my work by coming to the same conclusion by hand. I don't know how to use this wolfram site but if you can show how to algebraically manipulate the 4500/r equation to yield the same result it would be very appreciated. $\endgroup$
    – Ryan
    Oct 4 '19 at 1:30
  • $\begingroup$ Just multiply both sides by $r$ to get $4500=r(r^2+h^2)^{1/2}$, then do what you did for the other equation. $\endgroup$
    – R. Burton
    Oct 4 '19 at 1:34

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