Slope of Tangent

**RogueDemon** · October 14th 2010, 11:50 AM

The gravitational potential energy of the MIR space station, in joules, at a distance of x kilometers from the centre of the Earth, is described by the equation
$E = -\frac{(5.38)(10)^{19}}{x}$ . Determine the slope of the tangent to the graph at the given value of x.

$x = (6.5)(10)^6 m$

My answer:

$x = (6.5)(10)^6 m = 6500 km$

$m = \frac{E(6500 + 0.001) - E(6500)}{0.001}$

$m = \frac{(-\frac{(5.38)(10)^{19}}{(6500 + 0.001)}) -(-\frac{(5.38)(10)^{19}}{(6500)})}{0.001}$

$m = 1.273(10)^{12}$

Is this correct?

**CSM** · October 14th 2010, 01:08 PM

Yes your anwers is correct.

Your formula can be rewritten as:
$E = -(5.38)(10)^{19}* \frac{1}{x}$
Which has derivative:
$E' = (5.38)(10)^{19}* \frac{1}{x^2}$

Proof see: Calculus with polynomials - Wikipedia, the free encyclopedia
With your x, this gives $1.27337278 \times 10^{12}$

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