Answer:
See below.
Explanation:
Calculus was first developed to solve two main math problems that algebra could not handle.
Slope of a Tangent at a Point on a Curve
The first problem. - From algebra, we calculate the slope of a line by m = (y2 - y1)/(x2 - x1). You need two points to do that - point P1 with coordinates x1 and y1, and point P2, with coordinates x2 and y2.
But what if you want to know the slope of the line that is tangent to the function y = x³ - 2x² +3 at the one point where x = 5 and y = 5³ - 2*5² + 3 = 78?
You know nothing about the line except that is is tangent to that curve at that point. And you don't have 2 points on the line to use the slope formula.
Algebra gives no way for us to find that line's slope, but Calculus does.
Area Between a Function Curve and the x -axis
The second problem. - Now suppose we have that same curve, y = x³ - 2x² +3, and we want to know the area between the curve and the x-axis between x = 1 and x =2, like the yellow-shaded region in the picture I attached. Algebra gives us some long and tedious ways that we can approximate that area, but no way to say we know an exact number for the area. Calculus does give us a way to do that, and not that hard, either.
Calculus has been expanded beyond the original two problems, of course, but the method of how those two problems are solved form the foundation of calculus.
Okay, but how is that useful?
For linear relations, slope is a constant. But a lot of things don't have a linear relation - the speed of falling objects or objects in orbit around the earth, the rate water drains from a swimming pool, and so on.
Calculus is used for calculations that involve processes where change doesn't occur at a constant rate. The thing that is changing can be the velocity of an object, how fast something is warming up, or the curve of a line on a shape I am deigning or studying.
Shapes, areas, and volumes are easy for the shapes you learn in high school geometry, but a lot of shapes in the real world are more complicated. If I want a computerized lathe to cut a curvy form out of metal, and I want that form to take up a certain volume (and thus only a certain weight of metal), I need to calculate the volume of this weird shape. If I gave the robot lathe equations to describe the shape, I can apply Calculus to those equations and get that volume.