echan527
Nov 10, 2013
Undergraduate / "Ok class. You may start."; Stanford Intellectual Vitality [2]
"Ok class. You may start."
I quickly turned the paper over and there was only 1 problem:
"Show that (Tan(x+y)=x) = -(x^2)/(x^(2)-1)"
As it was a test over derivatives, I knew I'd have to apply some sort of derivative rule to solve this beast. As I couldn't isolate y, I decided that implicit differentiation would be the best approach. But when I reached what seemed like a dead end in the form of -sin(x+y), I almost gave up. I couldn't simplify any further. 5 min left. I started to worry. Then it hit me.
"What if I draw a triangle?"
Since tangent is opposite over adjacent, the opposite side must equal x and the adjacent is 1. The hypotenuse would then be √(x^2 + 1). Sin(x+y) would then equal x/√(x^2 + 1). Squaring the top and bottom would result in x^(2)/(x^2 + 1). Then add a negative sign because I had negative sin, and there lay my answer. I quickly wrote down the answer and submitted my paper.
The A was a nice gift, but my true reward was the application of other disciplines of mathematics to help reach what seemed like an improbable solution. Never would I have thought that a quick lesson in Pre-Calculus would save my butt and propel me to the top of my class. Our teacher said to think outside the box; how about outside the triangle?
Feedback and thought are very helpful, as English really isn't my strongest subject! Please don't be afraid to be critical as well!
"Ok class. You may start."
I quickly turned the paper over and there was only 1 problem:
"Show that (Tan(x+y)=x) = -(x^2)/(x^(2)-1)"
As it was a test over derivatives, I knew I'd have to apply some sort of derivative rule to solve this beast. As I couldn't isolate y, I decided that implicit differentiation would be the best approach. But when I reached what seemed like a dead end in the form of -sin(x+y), I almost gave up. I couldn't simplify any further. 5 min left. I started to worry. Then it hit me.
"What if I draw a triangle?"
Since tangent is opposite over adjacent, the opposite side must equal x and the adjacent is 1. The hypotenuse would then be √(x^2 + 1). Sin(x+y) would then equal x/√(x^2 + 1). Squaring the top and bottom would result in x^(2)/(x^2 + 1). Then add a negative sign because I had negative sin, and there lay my answer. I quickly wrote down the answer and submitted my paper.
The A was a nice gift, but my true reward was the application of other disciplines of mathematics to help reach what seemed like an improbable solution. Never would I have thought that a quick lesson in Pre-Calculus would save my butt and propel me to the top of my class. Our teacher said to think outside the box; how about outside the triangle?
Feedback and thought are very helpful, as English really isn't my strongest subject! Please don't be afraid to be critical as well!