Slope of Tangent

**madman1611** · July 18th 2007, 07:59 PM

Hey, guys I'm miserably stuck on this problem and require urgent help. Thanks.. Find the slope of the tangent to each curve at the given point. y=SQUARE ROOT 25-x^2 ; at points (3,4). Please provide a clear explanation as I am math noob, lol.

**Jhevon** · July 18th 2007, 08:11 PM

Originally Posted by madman1611

Hey, guys I'm miserably stuck on this problem and require urgent help. Thanks.. Find the slope of the tangent to each curve at the given point. y=SQUARE ROOT 25-x^2 ; at points (3,4). Please provide a clear explanation as I am math noob, lol.

the derivative gives the slope of the tangent line at any point. so all we need to do here, is find the derivative and plug in x = 3, since that is the x-value of the point we are concerned with

we have a composite function here, therefore we need the chain rule to differentiate this guy.

to refresh your memory, the chain rule says:

$\left[ f \left( g(x) \right) \right]' = f' \left( g(x) \right) \cdot g'(x)$

Now on to your problem:

$y = \sqrt {25 - x^2}$

$\Rightarrow y = \left( 25 - x^2 \right)^{ \frac {1}{2}}$

$\Rightarrow y' = \frac {1}{2} \left( 25 - x^2 \right)^{- \frac {1}{2}} \cdot (-2x)$

$\Rightarrow y' = -x \left( 25 - x^2 \right)^{ - \frac {1}{2}}$

or

$y' = - \frac {x}{ \sqrt {25 - x^2}}$

I leave the rest to you

**madman1611** · July 18th 2007, 08:23 PM

Hey, thanks bro. This is what I used to work around this:

y= SQUARE ROOT 25-x^2; at points (3,4)

m= lim h->0 f(a+h)-f(a)
___________
h

but my answer keeps turning out to be positive 3/4 and not negative.

**Jhevon** · July 18th 2007, 08:46 PM

Originally Posted by madman1611

Hey, thanks bro. This is what I used to work around this:

y= SQUARE ROOT 25-x^2; at points (3,4)

m= lim h->0 f(a+h)-f(a)
___________
h

but my answer keeps turning out to be positive 3/4 and not negative.

oh, you have to use the limit definition? the answer still turns out to be the same, which means you made a mistake somewhere

$f(x) = \sqrt {25 - x^2}$

Since, $f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$

We have:

$f'(3) = \lim_{h \to 0} \frac {\sqrt {25 - (3 + h)^2} - \sqrt {25 - 3^2}}{h}$

Rationalize the numerator

$= \lim_{h \to 0} \frac {25 - (3 + h)^2 - 25 + 3^2}{h \left( \sqrt {25 - (3 + h)^2} + \sqrt {25 - 3^2} \right)}$

expand the brackets in the numerator

$= \lim_{h \to 0} \frac {-9 - 6h - h^2 + 9}{h \left( \sqrt {25 - (3 + h)^2} + 4 \right)}$

the 9's cancel, now factor out an h from the top to cancel with the one in the bottom

$= \lim_{h \to 0} \frac {-h - 6}{\sqrt {25 - (3 + h)^2} + 4}$

Now take the limit

$= \frac {-6}{4 + 4}$

$= - \frac {3}{4}$

**madman1611** · July 18th 2007, 09:09 PM

I'm a moron! Similar question but my answer turns out to be positive. Can you please help with a little explanation:

Find the slope of the tangent to each curve at the given point.

y= 8 at x=5
----
Square root x+11

**Jhevon** · July 19th 2007, 04:31 AM

Originally Posted by madman1611

I'm a moron! Similar question but my answer turns out to be positive. Can you please help with a little explanation:

Find the slope of the tangent to each curve at the given point.

y= 8 at x=5
----
Square root x+11

Clarify. is this the question? $y = \frac {8}{\sqrt {x + 11}}$ ?

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