# high school / calculus

• January 30th 2007, 06:44 PM
philipsk
high school / calculus
I have received help from many of you last time i requested help and i thank you all for helping me out and not making me look like a fool in front of my child :) but now he needs my help again and he's giving me a few problems to show him how to solve so once again i'm seeking your help... if you guys could please show me how to solve these problems so i could try and teach it to him after teaching it to myself that would be great thank you all in advance sorry if i mistyped anything i'm doing this really fast because i have to go thank you all

http://img252.imageshack.us/img252/8803/calculushw5.jpg
• February 1st 2007, 01:02 PM
earboth
Quote:

Originally Posted by philipsk
...but now he needs my help again and he's giving me a few problems to show him how to solve...

http://img252.imageshack.us/img252/8803/calculushw5.jpg

Hello,

to 1.

To calculate the slope of a tangent you need the gradient of the function. The 1rst derivative gives you the gradient for any value of x. Use quotient rule and chain rule:
$f(x)=\frac{x}{\sqrt{2x-1}},\ \ x>\frac{1}{2}$
$f'(x)=\frac{1 \cdot \sqrt{2x-1}-x \cdot \frac{1}{2}(2x-1)^{-\frac{1}{2}} \cdot 2}{2x-1},\ \ x>\frac{1}{2}$

$f'(x)=\frac{1 \cdot \sqrt{2x-1}-x \cdot \frac{1}{2}(2x-1)^{-\frac{1}{2}} \cdot 2}{2x-1},\ \ x>\frac{1}{2}$. Simplify to

$f'(x)=\frac{x-1}{(\sqrt{2x-1})^3}$. You are looking for a horizontal tangent which have the slope zero:

f'(x) = 0 if x - 1 = 0. Thus x = 1

Plug in this value into f(x) and you'll get $f(1)=\frac{1}{\sqrt{2-1}}=1$

Thus the equation of the tangent is: y = 1

I've attached a diagram.

EB
• February 1st 2007, 09:58 PM
philipsk
thank you very much!

for the 2nd question do i take the derivative of the function then plug in x into the equation to find the equation of the line that intercepts at that point ?
• February 1st 2007, 11:59 PM
CaptainBlack
Quote:

Originally Posted by philipsk
thank you very much!

for the 2nd question do i take the derivative of the function then plug in x into the equation to find the equation of the line that intercepts at that point ?

The equation of the tangent will be of the general form of any straight line: $y=m\,x+c$, where $m$ is the slope of the line.

The derivative of the given graph will give you the slope of the tangent at the given point. you then have to find $c$ such that the line goes through the given point.

$f(x)=\sqrt{2x-1}$

so:

$\frac{df}{dx}=\frac{1}{2}\,\frac{1}{\sqrt{2x-1}}\,2=\frac{1}{\sqrt{2x-1}}$.

Now the point in question has $x$ coordinate $5$, so the slope of the tangent at the point in question is:

$m=\frac{1}{\sqrt{2\times 5-1}}=1/3$

So the equation of the tangent is:

$y=x/3+c$

and passes through $(5,3)$, so we must have:

$3=5/3+c$,

so $c=4/3$ and the equation of the tangent is:

$y=x/3+4/3$.

RonL
• February 2nd 2007, 12:33 AM
machi4velli
4.

a)Use the Chain rule
y = sin(x^2 + 1)^3
y' = 3(sin(x^2 + 1)^2)(cos(x^2+1)(2x)
y' = 6xsin(x^2+1)^2cos(x^2+1)

b)Use the Product and Chain rules.
f(x) = ((2x-3)^3)((x^2+1)^2)
f'(x) = (3)(2x-3)^2(2)(x^2+1)^2 + (2x-3)^3(2)(x^2+1)(2x)
f'(x) = 6(2x-3)^2(x^2+1)^2 + 4x(2x-3)^3(x^2+1)
f'(x) = 2(x^2+1)(2x-3)^2(3(x^2+1)+2x(2x-3))
f'(x) = 2(x^2+1)(2x-3)^2(7x^2-6x+3)

c)Use the Product and Chain rules.
f(x) = x(sinx)^2
f'(x) = (2x)(sin x)(cos x) + (sinx)^2

d)Use the Product and Chain rules
f(x) = x^2(x^2 + 4)^(1/2)
f'(x) = 2x(x^2+4)^(1/2) + x^2(2x)(1/2)(x^2+4)^(-1/2)
f'(x) = 2x(x^2+4)^(1/2) + x^3/(x^2+4)^(1/2)
• February 2nd 2007, 02:45 AM
earboth
Quote:

Originally Posted by philipsk
...but now he needs ... help again ...

Hello,

to #4:

with this problem I'm going to use implicite derivation:

I use $f(x,y)=x^2 y^2 - 9x^2 - 4y^2$. The derivation is calculated by

$f'(x,y)= - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$. That means here:

$f'(x,y)= - \frac{2x y^2-18x}{2x^2 y - 8y}$. Now plug in the values of x and y:

$f'(-4, 2\sqrt{3})= - \frac{-8 \cdot 12 + 72}{64 \sqrt{3} - 16 \sqrt{3}}=\frac{24}{48\sqrt{3}}=\frac{1}{6}\sqrt{3 }$

Now you can use the point-slope-formula of a straight line to calculate the equation of the tangent:

$\frac{y - 2\sqrt{3}}{x+4}=\frac{1}{6}\sqrt{3} \Longleftrightarrow y =\left(\frac{1}{6}\sqrt{3} \right) x+\frac{8}{3}\sqrt{3}$

EB
• February 2nd 2007, 06:49 AM
earboth
Quote:

Originally Posted by philipsk
...but now he needs my help again and he's giving me a few problems to show him how to solve so once again i'm seeking your help...

Hello,

it's me again.
If you are not familiar with implcite derivation I can now show you a different way to do #4.

First transform the given equation:

$x^2 y^2-9x^2-4y^2=0 \Longleftrightarrow y^2(x^2-4)=9x^2 \Longleftrightarrow y^2 = \frac{9x^2}{x^2-4}$ $\Longrightarrow y = \pm \sqrt{\frac{9x^2}{x^2-4}}$

The point in question lays above the x-axis (it's y-value is positive), therefore you have to use the positive square-root:

$y=f(x)= \sqrt{\frac{9x^2}{x^2-4}}$

Now calculate the 1rst derivative of this function (use chain rule, quotient rule):

$f'(x)=\frac{1}{2}\left( \frac{9x^2}{x^2-4} \right)^{-\frac{1}{2}}\cdot \frac{(x^2-4) \cdot 18x-9x^2 \cdot 2x}{(x^2-4)^2}$ $=\left( \frac{9x^2}{x^2-4} \right)^{-\frac{1}{2}}\cdot \frac{(x^2-4) \cdot 9x-9x^2\cdot x}{(x^2-4)^2}$ finally:

$f'(x)=\frac{-36x}{(x^2-4)^2 \cdot \sqrt{\frac{9x^2}{x^2-4}}}$

Now plug in the value x = -4 :

$f'(-4)=\frac{-36 \cdot (-4)}{((-4)^2-4)^2 \cdot \sqrt{\frac{9 \cdot (-4)^2}{(-4)^2-4}}}$ $=\frac{144}{144 \cdot \sqrt{12}}=\frac{1}{\sqrt{12}}=\frac{1}{2\sqrt{3}} =\frac{1}{6}\sqrt{3}$

To calculate the equation of the tangent have a look at my previous post.

EB
• February 2nd 2007, 12:43 PM
ticbol
Play more.
If you like to play more on the (x^2)(y^2) -9x^2 -4y^2 = 0, here is another way.

(x^2)(y^2) -9x^2 -4y^2 = 0
(x^2)[2yy'] +(y^2)[2x] -18x -8yy' = 0
Divide all by 2,
(x^2)yy' +xy^2 -9x -4yy' = 0
yy'(x^2 -4) +x(y^2 -9) = 0
yy'(x^2 -4) = -x(y^2 -9)
y' = -x(y^2 -9) / y(x^2 -4) ---------slope of tangent line anywhere on the curve.

At point (-4, 2sqrt(3)),
y' = -(-4)[(2sqrt(3))^2 -9] / 2sqrt(3)*[(-4)^2 -4]
y' = 4[3] / 2sqrt(3)*[12]
y' = 1 / 2sqrt(3)
y' = 1*sqrt(3) / 2*(sqrt(3))^2
y' = (sqrt(3)) /6 -------------------***

For the equation of the tangent line, using the point-slope form,
y -y1 = m(x -x1)
y - 2sqrt(3) = [sqrt(3) /6](x -(-4))
y -2sqrt(3) = [sqrt(3) /6]x +4[sqrt(3) /6]
y = [sqrt(3) /6]x +(2/3)sqrt(3) +2sqrt(3)
y = [sqrt(3) /6]x +(8/3)sqrt(3) -------------------the tangent line.