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.
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
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
$\displaystyle = \frac {-6}{4 + 4}$
$\displaystyle = - \frac {3}{4}$
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
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Square root x+11