Today (November 11) in class we went over the question from yesterday, number 5 on the worksheet called "Chain Rule combined with other differentiation rules." We learned that a circle is really two different explicit functions, one negative and one positive. The implicit function is x^2 + y^2 = 25.
We then did four review problems of composite functions. We reviewed how to find the derivative of tan(f(x)), sec(g(x)), e^(h(x)), and (f(x))^N. Then we started learning Implicit Differentiation. Using the slow method, we found the derivative of x^2 + (f(x))^2 =25, which is -x/f(x). The faster method involved using "y" instead of f(x), which meant that instead of using f'(x) we used dy/dx. The slow and fast methods are the same: they just use different notation.
We did a second example on the board and then we completed a worksheet in class called "Implicit Differentiation practice." We discussed that since you write the derivative of y^3 as 3y^2*(dy/dx), you can also write the derivative of x^3 as 3x^2* (dx/dx). We also said Newton would be interested in doing calculus for ellipses because the planets' orbits are ellipses, and he was fascinated with the planets.
On Friday November 14, we started class by getting our quizzed from the previous day back. After discussing the tough ones and answering questions, we went on to a new concept involving ripples on a pond. It started with the question of, if the radius of the ripple increased constantly, why doesn't the area increase in the same fashion? Before we answered this, we looked at Leibniz Notation. dy/dx is the change in y as a function of x. So, the change in area of the circle would be dA/dt. (Change in area as a function of time). So the question was posed, if the ripple's radius increase 3 cm/sec, how quickly is the area of the ripple increasing at time=3? So basically, we wanted to find dA/dt at 3. We already knew that dR/dt = 3 (Change in Radius as a function of time) so we needed a relationship between Radius and Area. Therefore, we took the equation of Area=(pi)r^2 and derived to find dA/dt= 2(pi)(r) (dR/dt). We then plugged in 3 for (dR/dt) and whatever radius was given to find (dA/dt). We did practice problems for the rest of class.
On November 10 we started to connect our knowledge of the chain rule with our other rules (product rule, quotient rule, etc.).
We started with an example that gave us y=e^(4/x^2) multiplied by x^3. In order to find the derivative of this function we had to look at the equation as a whole. Clearly, it involved the product rule, because we were using multiplication. As we used the product rule we saw that
y'= e^(4/x^2) x 3x^2 + x^3 x (the derivative of e^(4/x^2)).
We realized that inside the product rule, we found the chain rule. In order to find the derivative of this section we used our knowledge of equations that have e^(f(x)), and said that the derivative of this would be e^(4/x^2) x d/dx (4/x^2), which is simplified to e^(4/x^2) x d/dx (-8x^-3)
In conclusion our final equation looked like this:
y'= e^(4/x^2) multiplied by 3x^2 + x^3 multiplied by (e^(4/x^2)(-8x^-3))
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Today (November 11) in class we went over the question from yesterday, number 5 on the worksheet called "Chain Rule combined with other differentiation rules." We learned that a circle is really two different explicit functions, one negative and one positive. The implicit function is x^2 + y^2 = 25.
We then did four review problems of composite functions. We reviewed how to find the derivative of tan(f(x)), sec(g(x)), e^(h(x)), and (f(x))^N. Then we started learning Implicit Differentiation. Using the slow method, we found the derivative of x^2 + (f(x))^2 =25, which is -x/f(x). The faster method involved using "y" instead of f(x), which meant that instead of using f'(x) we used dy/dx. The slow and fast methods are the same: they just use different notation.
We did a second example on the board and then we completed a worksheet in class called "Implicit Differentiation practice." We discussed that since you write the derivative of y^3 as 3y^2*(dy/dx), you can also write the derivative of x^3 as 3x^2* (dx/dx). We also said Newton would be interested in doing calculus for ellipses because the planets' orbits are ellipses, and he was fascinated with the planets.
On Friday November 14, we started class by getting our quizzed from the previous day back. After discussing the tough ones and answering questions, we went on to a new concept involving ripples on a pond. It started with the question of, if the radius of the ripple increased constantly, why doesn't the area increase in the same fashion? Before we answered this, we looked at Leibniz Notation. dy/dx is the change in y as a function of x. So, the change in area of the circle would be dA/dt. (Change in area as a function of time). So the question was posed, if the ripple's radius increase 3 cm/sec, how quickly is the area of the ripple increasing at time=3? So basically, we wanted to find dA/dt at 3. We already knew that dR/dt = 3 (Change in Radius as a function of time) so we needed a relationship between Radius and Area. Therefore, we took the equation of Area=(pi)r^2 and derived to find dA/dt= 2(pi)(r) (dR/dt). We then plugged in 3 for (dR/dt) and whatever radius was given to find (dA/dt). We did practice problems for the rest of class.
On November 10 we started to connect our knowledge of the chain rule with our other rules (product rule, quotient rule, etc.).
We started with an example that gave us y=e^(4/x^2) multiplied by x^3. In order to find the derivative of this function we had to look at the equation as a whole. Clearly, it involved the product rule, because we were using multiplication. As we used the product rule we saw that
y'= e^(4/x^2) x 3x^2 + x^3 x (the derivative of e^(4/x^2)).
We realized that inside the product rule, we found the chain rule. In order to find the derivative of this section we used our knowledge of equations that have e^(f(x)), and said that the derivative of this would be e^(4/x^2) x d/dx (4/x^2), which is simplified to e^(4/x^2) x d/dx (-8x^-3)
In conclusion our final equation looked like this:
y'= e^(4/x^2) multiplied by 3x^2 + x^3 multiplied by (e^(4/x^2)(-8x^-3))
This simplifies to
y'= e^(4/x^2) (3x^2-8)
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