# Homework Prob

• Mar 13th 2006, 03:12 PM
tnkfub
Homework Prob
Find the equation of f the tangent to the curve at the given point:

y= (2x)/(x-1), point (2,4)

Also A canister is dropped from a helicopter hovering 50 m above the ground Unfortuanately its parachute does not open. Its been designed to withstand an impact velovity of 100 m / s Will it burst or not ? Acceleration = 9.8m/s^2
• Mar 13th 2006, 03:41 PM
topsquark
Quote:

Originally Posted by tnkfub
Find the equation of f the tangent to the curve at the given point:
y= (2x)/(x-1), point (2,4)

The slope of the tangent line is the first derivative. So:
$y=\frac{2x}{x-1}$

$y'=\frac{2x}{x-1}+\frac{2x*-1}{(x-1)^2}$
(Note: If this doesn't look like the quotient rule, that's because it isn't. I rarely use the quotient rule because I can't ever remember the correct form. I always move the denominator up and use the product rule.)

$y'=\frac{2x(x-1)-2x}{(x-1)^2}$

$y'=\frac{2x^2-4x}{(x-1)^2}$.

Now, to find the slope at the point (2,4) we plug in the x value:
$y'(2)=\frac{2*2^2-4*2}{(2-1)^2}=0$.

The equation for a line is y = mx + b. So we want the equation for the line with a slope of 0 that passes through the point (2,4). Thus:
$4=0*2+b$
$b=4$

Thus your line is y = 4.

-Dan
• Mar 13th 2006, 03:53 PM
topsquark
Quote:

Originally Posted by tnkfub
Also A canister is dropped from a helicopter hovering 50 m above the ground Unfortuanately its parachute does not open. Its been designed to withstand an impact velovity of 100 m / s Will it burst or not ? Acceleration = 9.8m/s^2

We know that motion under constant acceleration will follow the form:
$y=y_0+v_0t+(1/2)at^2$

In this case a = -9.8 m/s^2, y0= 50 m, and since it is dropped from a (presumably) stationary helicopter, v0 = 0 m/s. So:
$y=50-4.9t^2$.

How fast is it moving when it strikes the ground? Well, when it hits the ground, we know that y = 0 m. We need to know when this happens before we can calculate how fast. So:
$0=50-4.9t^2$
$t=\sqrt{\frac{50}{4.9}}$

Let's leave that as it is for now.

To find out how fast it is moving at that time, recall that $v=y'$ so:
$y=50-4.9t^2$
$v=y'=-9.8t$ (The negative sign simply means the object is moving downward.)

Plugging in the value of t when the object strikes the ground:
$v=-9.8*\sqrt{\frac{50}{4.9}}$

That's the exact number. The approximate value of v at the ground is about -31.3 m/s. Since the canister can take up to 100 m/s, the canister will not burst.

-Dan

Note: If you want to use Physics :), there is an equation that rarely comes up in a Math class that solves the problem immediately:
$v^2=v_0^2+2a(y-y_0)$. (For constant a only!)
You can use this to verify your solution if you like.
• Mar 13th 2006, 04:31 PM
tnkfub
Thanks ALot for the help Dan :D
• Mar 14th 2006, 03:41 AM
topsquark
You are welcome!

I live to serve... :D

-Dan