# Thread: Slope of Tangent

1. ## Slope of Tangent

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.

2. 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:

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

Now on to your problem:

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

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

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

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

or

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

I leave the rest to you

3. 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.

4. 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

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

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

We have:

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

Rationalize the numerator

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

expand the brackets in the numerator

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

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

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

Now take the limit

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

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

5. 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

6. 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? $\displaystyle y = \frac {8}{\sqrt {x + 11}}$?