- #1

gigi9

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also please give me an example that use the definite integral to compute accumulated changes.

Thanks alot

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- Thread starter gigi9
- Start date

- #1

gigi9

- 40

- 0

also please give me an example that use the definite integral to compute accumulated changes.

Thanks alot

- #2

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

e

differentiate both sides WRT x

e

x (ln x)' = 1

(ln x)' = 1 / x

so the derivative of ln x is 1/x

A relation is (to slightly modify the definition to make it easier to understand) simply a mapping from pairs of numbers to "true" or "false".

For instance:

x < y

is a relation.

Also, for any function f:

f(x) = y

is a relation. In fact,

f(x) = y and f(x) = z

can both be true only if y = z.

An example of a relation where one might use implicit differentiation is to find the slope of a point on the unit circle.

x

is a relation. Notice that y cannot be written as a

2 x + 2 y (dy/dx) = 0

2 y (dy/dx) = - 2x

(dy/dx) = -x / y

So, the slope of the tangent line to any point (x, y) of the unit circle is (dy/dx) = -x / y

Thanks

also please give me an example that use the definite integral to compute accumulated changes.

Suppose I throw a ball into the air straight up with a speed of 19.6 meters per second. How high is it after 2 seconds?

Well, acceleration is simply the rate change of velocity, so we can find the velocity at any time t with a definite integral:

v(t) - v(0) = &int

v(t) - 19.6 = &int

v(t) = 19.6 + (-9.8) * (t - 0)

v(t) = 19.6 - 9.8 t

Now, velocity is simply the rate chance of position, so we can find the height at time 2 with a definite integral:

x(2) - x(0) = &int

x(2) - 0 = &int

x(2) = [19.6 s - 4.9 s

x(2) = (19.6 * 2 - 4.9 * 2

x(2) = 19.6

So after 2 seconds, the ball is 19.6 meters high.

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