How is calculus different from algebra?
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12 Answers
Mark Eichenlaub, physics curriculum developer at Art of Problem Solving
Originally Answered: How is calculus different than algebra?
Once you learn the basics of calculus, solving calculus problems will be almost indistinguishable from solving algebra problems.
For example, in algebra you learn to distribute multiplication, so
[math] a(b+c) = ab + ac[/math]
In calculus, you will learn the product rule for derivatives, so
[math] \frac{\mathrm{d}}{\mathrm{d}x}(fg) = f\frac{\mathrm{d}}{\mathrm{d}x}g + g \frac{\mathrm{d}}{\mathrm{d}x}f[/math]
Even if you don't know what is meant by [math]\frac{\mathrm{d}}{\mathrm{d}x}[/math], you can still see that the formula is not very different from algebra formulas, and solving calculus problems feels very similar to solving algebra problems. So the details of carrying out the computation are almost the same.
Here are some differences in terms of the concepts and uses.
Calculus deals with limits.
Suppose you want to know the equation for a line that is tangent to a circle.
To solve this problem with algebra, you could start by writing the equation for a circle
[math] (x-a)^2 + (y-b)^2 = r^2[/math]
and the equation for a line
[math] y = mx + c [/math]
Then you draw a picture like this:
What's special about a tangent line is that it contacts the circle in exactly one point. Most lines will either miss the circle entirely or cross it in two points.
What's that have to do with algebra? The intersection points are where the two equations are both true, so you can substitute the second one into the first one.
If you do this, you'll see you have a quadratic equation for [math]x[/math]. A quadratic equation usually has zero or two solutions, but it can have exactly one solution if the discriminant is zero. By setting the discriminant equal to zero you find the condition on [math]a, b, m, c[/math] such that the line is tangent to the circle.
To solve the problem with calculus, you'd consider a series of secant lines approximating a tangent.
The thick line is the true tangent. The thin lines get closer and closer to it. We can find equations for the thin lines because we know two different points on them. We use the pattern in these equations to figure out the equation of the thick line.
In order to master this problem with calculus, you need to learn how to deal with things getting closer and closer to some true value.
(I made this last image with Geogebra, a great tool for learning about the connections between algebra, calculus, and geometry at the high school level.
http://www.geogebra.org/cms/)
Calculus is more useful to model physical processes.
It seems like a simple problem, but it took people a long time to figure out the rule for how things fall in the absence of air resistance. The rule is
[math]x = \frac{1}{2}g t^2[/math]
[math]x[/math] is the distance the object falls, [math]g[/math] is the strength of gravity, and [math]t[/math] is how long the object falls. With algebra, you can now solve how long it will take to fall any given distance. But what if you want to know how fast something will be going when it hits the ground?
You might be able to figure it out if you're very clever, but you'd essentially wind up using the tools of calculus to do it. Otherwise you'd need someone to give you a new formula
[math]\frac{1}{2} v^2 = gx [/math]
and if you want a slightly different problem on the same stuff, you'd need to memorize more and more algebraic formulas. Sadly, this is still how introductory physics is commonly taught in the US, even in college.
Once you know calculus, all you need to know is
[math] \frac{\mathrm{d^2}}{\mathrm{d}t^2}x = g[/math]
and you can easily derive all the formulas related to falling bodies. (Instead of this one calculus formula, it is common to list five algebraic ones and have students memorize all five.)
Further, what if the object falls from a very high distance so that gravity is changing during the fall? What if there's air drag proportional to the velocity, or to the square of the velocity? What if the object has a thruster? What if there's a cross-wind? Calculus lets you deal with all these situations.
In the previous problem, we saw that calculus lets you take a series of lines that have smaller and smaller separation from each other and make sense of the situation. In physics, calculus lets you take moments in time that have smaller and smaller separation (one second, one tenth of a second, one hundredth of a second...) and make sense of it.
Things like the effect of wind change rapidly from moment to moment, so initially you only understand one little moment worth of them. Calculus lets you build up those tiny understandings into figuring out what will happen in the long run.
Calculus and algebra are closely related.
You will constantly use algebra while doing calculus. In fact, I usually don't even notice when I switch from doing algebra to doing calculus. It is all just "doing a calculation." (Unless it's something like a tricky calculus integral and I have to stop and think for a while.)
A mathematician could speak to the interconnections of the subjects better than I, so I'll bow out to avoid overstepping myself, but after an initial period of adjustment, you should find calculus to be very familiar if you are already comfortable with algebra. Calculus may let you understand some topics in algebra better, and algebra will be essential to doing calculus.
For example, in algebra you learn to distribute multiplication, so
[math] a(b+c) = ab + ac[/math]
In calculus, you will learn the product rule for derivatives, so
[math] \frac{\mathrm{d}}{\mathrm{d}x}(fg) = f\frac{\mathrm{d}}{\mathrm{d}x}g + g \frac{\mathrm{d}}{\mathrm{d}x}f[/math]
Even if you don't know what is meant by [math]\frac{\mathrm{d}}{\mathrm{d}x}[/math], you can still see that the formula is not very different from algebra formulas, and solving calculus problems feels very similar to solving algebra problems. So the details of carrying out the computation are almost the same.
Here are some differences in terms of the concepts and uses.
Calculus deals with limits.
Suppose you want to know the equation for a line that is tangent to a circle.
To solve this problem with algebra, you could start by writing the equation for a circle
[math] (x-a)^2 + (y-b)^2 = r^2[/math]
and the equation for a line
[math] y = mx + c [/math]
Then you draw a picture like this:
What's special about a tangent line is that it contacts the circle in exactly one point. Most lines will either miss the circle entirely or cross it in two points.
What's that have to do with algebra? The intersection points are where the two equations are both true, so you can substitute the second one into the first one.
If you do this, you'll see you have a quadratic equation for [math]x[/math]. A quadratic equation usually has zero or two solutions, but it can have exactly one solution if the discriminant is zero. By setting the discriminant equal to zero you find the condition on [math]a, b, m, c[/math] such that the line is tangent to the circle.
To solve the problem with calculus, you'd consider a series of secant lines approximating a tangent.
The thick line is the true tangent. The thin lines get closer and closer to it. We can find equations for the thin lines because we know two different points on them. We use the pattern in these equations to figure out the equation of the thick line.
In order to master this problem with calculus, you need to learn how to deal with things getting closer and closer to some true value.
(I made this last image with Geogebra, a great tool for learning about the connections between algebra, calculus, and geometry at the high school level.
http://www.geogebra.org/cms/)
Calculus is more useful to model physical processes.
It seems like a simple problem, but it took people a long time to figure out the rule for how things fall in the absence of air resistance. The rule is
[math]x = \frac{1}{2}g t^2[/math]
[math]x[/math] is the distance the object falls, [math]g[/math] is the strength of gravity, and [math]t[/math] is how long the object falls. With algebra, you can now solve how long it will take to fall any given distance. But what if you want to know how fast something will be going when it hits the ground?
You might be able to figure it out if you're very clever, but you'd essentially wind up using the tools of calculus to do it. Otherwise you'd need someone to give you a new formula
[math]\frac{1}{2} v^2 = gx [/math]
and if you want a slightly different problem on the same stuff, you'd need to memorize more and more algebraic formulas. Sadly, this is still how introductory physics is commonly taught in the US, even in college.
Once you know calculus, all you need to know is
[math] \frac{\mathrm{d^2}}{\mathrm{d}t^2}x = g[/math]
and you can easily derive all the formulas related to falling bodies. (Instead of this one calculus formula, it is common to list five algebraic ones and have students memorize all five.)
Further, what if the object falls from a very high distance so that gravity is changing during the fall? What if there's air drag proportional to the velocity, or to the square of the velocity? What if the object has a thruster? What if there's a cross-wind? Calculus lets you deal with all these situations.
In the previous problem, we saw that calculus lets you take a series of lines that have smaller and smaller separation from each other and make sense of the situation. In physics, calculus lets you take moments in time that have smaller and smaller separation (one second, one tenth of a second, one hundredth of a second...) and make sense of it.
Things like the effect of wind change rapidly from moment to moment, so initially you only understand one little moment worth of them. Calculus lets you build up those tiny understandings into figuring out what will happen in the long run.
Calculus and algebra are closely related.
You will constantly use algebra while doing calculus. In fact, I usually don't even notice when I switch from doing algebra to doing calculus. It is all just "doing a calculation." (Unless it's something like a tricky calculus integral and I have to stop and think for a while.)
A mathematician could speak to the interconnections of the subjects better than I, so I'll bow out to avoid overstepping myself, but after an initial period of adjustment, you should find calculus to be very familiar if you are already comfortable with algebra. Calculus may let you understand some topics in algebra better, and algebra will be essential to doing calculus.
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हिन्दी में: कैलकुलस और अलजेब्रा में क्या फर्क है?