Name: MAT183 CALCULUS I
Uploaded: 2022-01-25
Duration: 3 min 50 s
Description: Application of Differentiation. Three paintbrushes with paint on the tips.

Application of Differentiation. Three paintbrushes with paint on the tips. 

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Three paintbrushes with paint on the tips

We are really grateful because we manage to complete this video presentation within the given time by our lecturer , Madam Liyana . This assignment is impossible to be done without the amazing cooperation abd effort of our group members Mirrah Syafiah , Alia Athirah , and Fazizatul Hana. We also sincerely thank our lecturer of Calculus 1, Madam Liyana for the guidance and willingness to help us in what we are lacking when completing this assignment and surely for teaching us this course. Last but not least, we would like to express our gratitude to our family members who keep encouraging us all the this time.. 

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We are really grateful because we manage to complete this video presentation within the given time by our lecturer , Madam  Liyana . This assignment is impossible to be done without the amazing cooperation  abd  effort of our group members  Mirrah   Syafiah , Alia  Athirah , and  Fazizatul  Hana. We also sincerely thank our lecturer of Calculus 1, Madam  Liyana  for the guidance and willingness to help us in what we are lacking when completing this assignment and surely for teaching us this course. Last but not least, we would like to express our gratitude to our family members who keep encouraging us all the this time.

Meet out team!. Presenter 1. Presenter 3. Name :. 

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Introduction. Differentiation is the process of finding the derivative or the rate of change of the function. The practical method of differentiation is simple, using knowledge of three basic derivatives, four operational rules, and how to operate a function. It can be done by algebraic operations.. 

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Differentiation is the process of finding the derivative or the rate of change of the function. The practical method of differentiation is simple, using knowledge of three basic derivatives, four operational rules, and how to operate a function. It can be done by algebraic operations.

The three basic  derivaties  are: •	Algebraic Function (D( xn ) =  nxn − 1 where n is a real number) •	Trigonometric Function (D(sin x) = cos x and D(cos x) = −sin x) •	Exponential Function (D(ex) = ex)

For the other operational rules, which is product rules. It provides a way to differentiate compound functions. The chain rule states that the derivative of a composite function is given by a product, as D(f(g(x))) = Df(g(x)). 

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For the other operational rules, which is product rules. It provides a way to differentiate compound functions. The chain rule states that the derivative of a composite function is given by a product, as D(f(g(x))) = Df(g(x))

For a function consisting of a combination of functions of these classes, the theory is to distinguish the sum, product, or quotient of any two functions f (x) and g (x) whose derivatives are known. It provides the following basic rules :

Quotient Rules (D(f/g) = ( gDf  −  fDg )/g2)

Problem Statement. A SPHERICAL BALLOON IS BEING INFLATED AT THE RATE OF 8 CM3/SEC. 

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A SPHERICAL BALLOON IS BEING INFLATED AT THE RATE OF 8 CM3/SEC

WHICH THE RADIUS OF THE BALLOON IS CHANGING WHEN THE RADIUS IS 10CM .

Problem Solution ;. dv/ dt =8 cm3 /sec ( dr )/ dt =? cm3 /sec. 

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dv/ dt   =8 cm3 /sec ( dr  )/ dt   =? cm3 /sec

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Conclusion. Based on the problem statement there is a spherical balloon being inflated at the rate of 8cm3 /sec.as we know the basic formula when we get any question about sphere are. 

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Based on the problem statement there is a spherical balloon being inflated at the rate of 8cm3 /sec.as we know the basic formula when we get any question about sphere are

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As we are asked to determine the rate at which the radius of the inflated balloon is changing when the radius is 10cm. 

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As we are asked to determine the rate at which the radius of the inflated balloon is changing when the radius is 10cm

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So we have to apply differentiation in our problem-solving. based on our problem-solving, we can see that the volume of the sphere change at a rate of 8cm3 /sec. Secondly, we have to differentiate the volume of the sphere formula to get the dv/ dr and it will be 4πr^2. after we have the dv/ dr we can inverse it to be dr /dv which we need to use in the formula to find the rate at which the radius of the balloon is changing when the radius is 10cm. 

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So we have to apply differentiation in our problem-solving. based on our problem-solving, we can see that the volume of the sphere change at a rate of 8cm3 /sec. Secondly, we have to differentiate the volume of the sphere formula to get the dv/ dr  and it will be 4πr^2. after we have the dv/ dr  we can inverse it to be  dr /dv which we need to use in the formula to find the rate at which the radius of the balloon is changing when the radius is 10cm

Lastly, we can solve the  dr /dt because we already got the answer for  dr /dv and dv/dt. the final answer will be 1/50π cm/sec. this shows that the radius of the sphere is changing at a rate 1/50π cm each second when the balloon is being inflated. So the radius will keep increasing with the rate of 1/50π cm per second when it is being inflated

A lamp shining light onto a desk. A cat. Thank You For Watching!. 