This speed translates to approximately 95 mph—a major-league fastball.
Surface area generated by a parametric curve
Recall the problem of finding the surface area of a volume of revolution. In
Curve Length and Surface Area , we derived a formula for finding the surface area of a volume generated by a function
$y=f\left(x\right)$ from
$x=a$ to
$x=b,$ revolved around the
x -axis:
We now consider a volume of revolution generated by revolving a parametrically defined curve
$x=x\left(t\right),y=y\left(t\right),a\le t\le b$ around the
x -axis as shown in the following figure.
The analogous formula for a parametrically defined curve is
The derivative of the parametrically defined curve
$x=x\left(t\right)$ and
$y=y\left(t\right)$ can be calculated using the formula
$\frac{dy}{dx}=\frac{{y}^{\prime}(t)}{{x}^{\prime}(t)}.$ Using the derivative, we can find the equation of a tangent line to a parametric curve.
The area between a parametric curve and the
x -axis can be determined by using the formula
$A={\displaystyle {\int}_{{t}_{1}}^{{t}_{2}}y\left(t\right){x}^{\prime}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt}.$
The arc length of a parametric curve can be calculated by using the formula
$s={\displaystyle {\int}_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt}.$
The surface area of a volume of revolution revolved around the
x -axis is given by
$S=2\pi {\displaystyle {\int}_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime}\left(t\right)\right)}^{2}+{\left({y}^{\prime}\left(t\right)\right)}^{2}}dt}.$ If the curve is revolved around the
y -axis, then the formula is
$S=2\pi {\displaystyle {\int}_{a}^{b}x\left(t\right)\sqrt{{\left({x}^{\prime}\left(t\right)\right)}^{2}+{\left({y}^{\prime}\left(t\right)\right)}^{2}}dt}.$
Key equations
Derivative of parametric equations $\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}=\frac{{y}^{\prime}\left(t\right)}{{x}^{\prime}\left(t\right)}$
Second-order derivative of parametric equations $\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\left(d\text{/}dt\right)\left(dy\text{/}dx\right)}{dx\text{/}dt}$
Area under a parametric curve $A={\displaystyle {\int}_{a}^{b}y\left(t\right){x}^{\prime}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt}$
Arc length of a parametric curve $s={\displaystyle {\int}_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt}$
Surface area generated by a parametric curve $S=2\pi {\displaystyle {\int}_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime}\left(t\right)\right)}^{2}+{\left({y}^{\prime}\left(t\right)\right)}^{2}}dt}$
For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?