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Find the equation of a tangent line to the curve of the function $\text{\hspace{0.17em}}f(x)=5{x}^{2}-x+4\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=2.$
$y=19x-16$
If a function measures position versus time, the derivative measures displacement versus time, or the speed of the object. A change in speed or direction relative to a change in time is known as velocity . The velocity at a given instant is known as instantaneous velocity .
In trying to find the speed or velocity of an object at a given instant, we seem to encounter a contradiction. We normally define speed as the distance traveled divided by the elapsed time. But in an instant, no distance is traveled, and no time elapses. How will we divide zero by zero? The use of a derivative solves this problem. A derivative allows us to say that even while the object’s velocity is constantly changing, it has a certain velocity at a given instant. That means that if the object traveled at that exact velocity for a unit of time, it would travel the specified distance.
Let the function $\text{\hspace{0.17em}}s\left(t\right)\text{\hspace{0.17em}}$ represent the position of an object at time $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ The instantaneous velocity or velocity of the object at time $\text{\hspace{0.17em}}t=a\text{\hspace{0.17em}}$ is given by
A ball is tossed upward from a height of 200 feet with an initial velocity of 36 ft/sec. If the height of the ball in feet after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by $\text{\hspace{0.17em}}s(t)=\mathrm{-16}{t}^{2}+36t+200,$ find the instantaneous velocity of the ball at $\text{\hspace{0.17em}}t=2.$
First, we must find the derivative $\text{\hspace{0.17em}}{s}^{\prime}\left(t\right)$ . Then we evaluate the derivative at $\text{\hspace{0.17em}}t=2,$ using $\text{\hspace{0.17em}}s\left(a+h\right)=-16{\left(a+h\right)}^{2}+36(a+h)+200\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}s\left(a\right)=-16{a}^{2}+36a+200.$
A fireworks rocket is shot upward out of a pit 12 ft below the ground at a velocity of 60 ft/sec. Its height in feet after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by $\text{\hspace{0.17em}}s=-16{t}^{2}+60t-12.\text{\hspace{0.17em}}$ What is its instantaneous velocity after 4 seconds?
–68 ft/sec, it is dropping back to Earth at a rate of 68 ft/s.
Access these online resources for additional instruction and practice with derivatives.
Visit this website for additional practice questions from Learningpod.
average rate of change | $\text{AROC}=\frac{f\left(a+h\right)-f\left(a\right)}{h}$ |
derivative of a function | ${f}^{\prime}(a)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(a+h\right)-f\left(a\right)}{h}$ |
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