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Interpretation of the second derivative : Now that we've discover some higher order derivatives we have to probably talk regarding an interpretation of the second derivative.
If the position of an object is specified by s(t) we know that the velocity is first derivative of the position.
v (t ) = s′ (t )
First derivative of any velocity is the acceleration of object; however since it is the first derivative of the position function we can also think of the acceleration as the second derivative of the position function.
a (t ) = v′ (t ) = s′′ (t )
Alternate Notation : There is couple of alternate notation for higher order derivatives. Recall as well that there was a fractional notation for the first derivative.
f ′ ( x ) = df /dx
We should extend this to higher order derivatives.
f ′′ ( x )= d 2 y / dx f ′′′ ( x ) = d 3 y/ dx etc.
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the sides of a right angle triangle are a,a+d,a+2d with a and d both positive.the ratio of a to d a)1:2 b)1:3 c)3:1 d)5:2 answer is (c) i.e. 3:1 Solution: Applying
I need to follow the pattern .125,.25,.375,.5, ?
Consider x € R. Then the magnitude of x is known as it's absolute value and in general, shown by |x| and is explained as Since the symbol always shows the nonnegative
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Substitution Rule ∫ f ( g ( x )) g′ ( x ) dx = ∫ f (u ) du, where, u = g ( x ) we can't do the following integrals through general rule. This looks considerably
For the initial value problem y' + 2y = 2 - e -4t , y(0) = 1 By using Euler's Method along with a step size of h = 0.1 to get approximate values of the solution at t = 0.1, 0
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