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
The slope of the tangent line is the first derivative. So:Originally Posted by tnkfub
(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.)
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Now, to find the slope at the point (2,4) we plug in the x value:
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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:
Thus your line is y = 4.
-Dan
We know that motion under constant acceleration will follow the form:
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:
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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:
Let's leave that as it is for now.
To find out how fast it is moving at that time, recall thatso:
(The negative sign simply means the object is moving downward.)
Plugging in the value of t when the object strikes the ground:
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:
. (For constant a only!)
You can use this to verify your solution if you like.
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