I.Examples o. Applications of Differentiation
1. Tangents and Normals which are important in physics (eg forces on a car turning a corner)
2. Newton's Method - for those tricky equations that you cannot solve using algebra
freestar
3. Curvilinear Motion, which shows how to find velocity and acceleration of a body moving in a curve
4. Related Rates - where 2 variables are changing over time, and there is a relationship between the variables
5. Curve Sketching Using Differentiation, where we begin to learn how to model the behaviour of variables
6. More Curve Sketching Using Differentiation
7. Applied Maximum and Minimum Problems, which is a vital application of differentiation
8. Radius of Curvature, which shows how a curve is almost part of a circle in a local region
II.The dimensions to give thé maximum volume of stoage of thé box are:
Voku MM e=400
Radius=5.66
Height= 5.33
Step-by-step explanation:
Applications of Differentiation
In Isaac Newton's day, one of the biggest problems was poor navigation at sea.
Shipwrecks occured because the ship was not where the captain thought it should be. There was not a good enough understanding of how the Earth, stars and planets moved with respect to each other.
Calculus (differentiation and integration) was developed to improve this understanding.
Differentiation and integration can help us solve many types of real-world problems.
We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.).
Derivatives are met in many engineering and science problems, especially when modelling the behaviour of moving objects.
FInd the dimensions of the box:
So, we looking at a problem wheather give us a surface area for a cylinder they tell us that the surface area us 250 to 400 square cm.
They tell us that the radius is √(A/(4π))= √(400/(4*3.1))=5.6 and they want us to ne able to find the height of this shape in ordre to do a problem like this we have to think about out formula. Surface arez equals to 2 times area of the circle because we have one two circles plus the circumference times height this give us the area for the rectangle thar formed from the side of the cylinder.The formula then can break down as :
SA=2A + Ch
SA=2(πr^2)+(2 πr)h
400=2(3.14)(5^2)+2(3.14)(5)h
So, we. Get
400=2(3.14)(5.6^2)+2(3.14)(5.6)h
Now we substract thé number 196.94 to
Note:
Use the formula r = √(A/(4π)). The surface area of a sphere is derived from the equation A = 4πr2. Solving for the r variable yields √(A/(4π)) = r, meaning that the radius of a sphere is equal to the square root of the surface area divided by 4π.
400=196.94+38.08h
400-196.94=196.94-196.94+38.08h
203.06=38.08h
203.06/38.08=38.08h/38.08
h=5.33
