# basic calculus problems

• Jan 18th 2007, 12:42 PM
philipsk
basic calculus problems
my son just asked me to help him with some calculus but i haven't done calculus in 30 years so i can't really help him... i was hoping that someone here could help. his homework is do tomorrow. so, time is essential.. well, i was hoping that someone could hopefully show me how to do these problems so i could show my son how to do them.

Thanks in advance to all that this may concern.

http://img259.imageshack.us/img259/1815/calculusno2.jpg
• Jan 18th 2007, 12:51 PM
ThePerfectHacker
Quote:

Originally Posted by philipsk
my son just asked me to help him with some calculus but i haven't done calculus in 30 years so i can't really help him...

Does he know that you are getting help or does he think that you know?

1)$\displaystyle f(x)=x-\sin 2x$
$\displaystyle f'(x)=(x)'-(\sin 2x)'=1-2\cos 2x$ (chain rule).

5)$\displaystyle y=\frac{\sin x}{4+\cos x}$
Quotient rule,
$\displaystyle y'=\frac{(\sin x)'(4+\cos x)-\sin x(4+\cos x)'}{(4+\cos x)^2}$
$\displaystyle y'=\frac{\cos x(4+\cos x)-\sin x(-\sin x)}{(4+\cos x)^2}$
$\displaystyle y'=\frac{4\cos x+\cos^2 x+\sin^2 x}{(4+\cos x)^2}=\frac{4\cos x+1}{(4+\cos x)^2}$
• Jan 18th 2007, 01:26 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Does he know that you are getting help or does he think that you know?

philipsk is a dad he is entitled to our help to appear omnicient in the eyes
of his chidren.

RonL
• Jan 18th 2007, 01:28 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
philipsk is a dad he is entitled to our help to appear omnicient in the eyes
of his chidren.

I should have added that he should not say that he is getting help to his son, I forgot to.
• Jan 18th 2007, 01:48 PM
CaptainBlack
2. Find the equation of the tangent to the curve $\displaystyle y=\frac{\sqrt{x}}{x+6}$ at $\displaystyle (4,0.2)$

The equation of a line (and the tangent is a line) is of the form $\displaystyle y=mx+c$, and this is a tangent at the given point on the curve if $\displaystyle m$ is equal to the slope or gradient of the curve at the point and c is chosen so that the line passes through the point.

Now the slope of the curve at a point is equal to the derivative of $\displaystyle y$ with respect to [tex][x/math] at the point. To evaluate this derivative we need the quotient rule:

$\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$,

with $\displaystyle f(x)=\sqrt{x}$ and $\displaystyle g(x)=x+6$, so $\displaystyle f'(x)=\frac{1}{2 \sqrt{x}}$, and $\displaystyle g'(x)=1$

Hence:

$\displaystyle \frac{dy}{dx}=\frac{\frac{x+6}{2\sqrt{x}}-\sqrt{x}}{(x+6)^2}$

Now we could simplify this but I think it better to just evaluate it as it is at $\displaystyle x=4$, which I make: $\displaystyle \left. \frac{dy}{dx}\right|_{x=4}=0.005$.

So our line has the form: $\displaystyle y=0.005\,x+c$. Now when $\displaystyle x=4\ y=0.2$, so: $\displaystyle 0.2=0.005\,4+c$, hence $\displaystyle c=0.18$, so finally we have the equation of the tangent at the given point is:

$\displaystyle y=0.005\,x+0.18$

RonL

(now I expect one of the other helpers will point out some dreadful mistake in my algebra:( )
• Jan 18th 2007, 01:56 PM
topsquark
2. Find the equation of the line tangent to $\displaystyle y = \frac{\sqrt{x}}{x+6}$ at the point (4, 0.2).

The equation of any line may be expressed as
$\displaystyle y = mx + b$
where m is the slope of the line and b is the y-intercept (0, b).

The slope of the line tangent to $\displaystyle y = \frac{\sqrt{x}}{x+6}$ at the point (x, y) is the first derivative of y with respect to x. So by the quotient rule:
$\displaystyle y' = \frac{\frac{1}{2} \cdot \frac{1}{\sqrt{x}} \cdot (x + 6) - \sqrt{x} \cdot (1)}{(x+6)^2}$ <-- Mulitply top and bottom by $\displaystyle 2\sqrt{x}$

$\displaystyle y' = \frac{x + 6 - 2x}{2 \sqrt{x} (x + 6)^2}$

$\displaystyle y' = \frac{-x + 6}{2 \sqrt{x} (x + 6)^2}$

So the slope of the tangent line to the curve at the point (4, 0.2) is:
$\displaystyle m = \frac{-4 + 6}{2 \sqrt{4} (4 + 6)^2} = 0.005$ <-- Or 1/200.

So the equation of the tangent line at (4, 0.2) is:
$\displaystyle y = 0.005x + b$

Now we need to find b, so put in the point coordinates:
$\displaystyle 0.2 = 0.005 \cdot 4 + b$

$\displaystyle b = 0.18$ <-- Or 9/50

Thus the equation of the line is:
$\displaystyle y = \frac{1}{200}x + \frac{9}{50}$

I have attached a graph of the function and the tangent line.

-Dan

Quote:

Originally Posted by CaptainBlack
(now I expect one of the other helpers will point out some dreadful mistake in my algebra:( )

At least we agree! :)
• Jan 18th 2007, 02:09 PM
topsquark
3. Find the equation of the line tangent to $\displaystyle y = x^8cos(x)$ at $\displaystyle ( \pi, -\pi^8)$.

Same method as before.

$\displaystyle y' = 8x^7cos(x) + x^8 \cdot - sin(x)$

$\displaystyle y' = -x^8sin(x) + 8x^7cos(x)$

At the point $\displaystyle ( \pi, -\pi^8)$ this is:
$\displaystyle m = -(\pi)^8sin(\pi) + 8(\pi)^7cos(\pi) = 0 - 8\pi^7 = -8\pi^7$

So we have the tangent line:
$\displaystyle y = (-8\pi^7)x + b$
at the point $\displaystyle ( \pi, -\pi^8)$. Again, we need b, so:

$\displaystyle -\pi^8 = (-8\pi^7) \cdot \pi + b$

$\displaystyle b = -\pi^8 + 8\pi^8 = 7\pi^8$

Thus the tangent line is:
$\displaystyle y = (-8\pi^7)x + 7\pi^8$

Again, I have attached a graph. (The scale of the y-axis hides some of the features of the graph.)

-Dan
• Jan 18th 2007, 02:31 PM
galactus
For #7, polynomials are among the easiest functions to differentiate. They're everywhere continuous and very well-behaved.

$\displaystyle \frac{d}{dx}[2x^{3}-3x^{2}-36x+4]=6x^{2}-6x-36=x^{2}-x-6$

Factor:

$\displaystyle (x-3)(x+2)$

As you can see, you have zeros at x=3 and x=-2

See the graph. Note the horizontal tangents as the maximum and minimum.