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Example. An object moving along a straight line has position given by $s(t) = \sqrt{t+3}$, where position is measured in meters and time in seconds. Find the (instantaneous) velocity of the object at time $t = 6$. Explanation. Velocity is the instantaneous rate of change of position with respect to time. We are being asked to find $v(6) = s'(6)$. We have to adjust the definition of the the derivative, namely, replace $x$ with $t$, $s'(6) = \lim_{t \to 6} \frac{s(t) - s(6)}{t - 6} = \lim_{t \to 6} \frac{\sqrt{t+3} - 3}{t - 6}$. Multiply by $\frac{\sqrt{t+3} + 3}{\sqrt{t+3} + 3}$, in order to rationalize the expression, Now simplify the numerator, Combine like-terms, $s'(6) = \lim_{t \to 6} \frac{(\sqrt{t+3} - 3)}{t - 6} (\frac{\sqrt{t+3} + 3}{\sqrt{t+3} + 3})$. $s'(6) = \lim_{t \to 6} \frac{t + 3 - 9}{(t - 6)(\sqrt{t+3} + 3)}$ $s'(6) = \lim_{t \to 6} \frac{t - 6}{(t - 6)(\sqrt{t+3} + 3)}$. Take the limit as $t$ goes to 6, $s'(6) = \lim_{t \to 6} \frac{1}{\sqrt{t+3} + 3} = \frac{1}{?}$ 

          Example. An object moving along a straight line has position given by $s(t) = \sqrt{t+3}$, where position is measured in meters and time in seconds. Find the
(instantaneous) velocity of the object at time $t = 6$.
Explanation. Velocity is the instantaneous rate of change of position with respect to time. We are being asked to find $v(6) = s'(6)$. We have to adjust the
definition of the the derivative, namely, replace $x$ with $t$,
$s'(6) = \lim_{t \to 6} \frac{s(t) - s(6)}{t - 6} = \lim_{t \to 6} \frac{\sqrt{t+3} - 3}{t - 6}$.
Multiply by $\frac{\sqrt{t+3} + 3}{\sqrt{t+3} + 3}$, in order to rationalize the expression,
Now simplify the numerator,
Combine like-terms,
$s'(6) = \lim_{t \to 6} \frac{(\sqrt{t+3} - 3)}{t - 6} (\frac{\sqrt{t+3} + 3}{\sqrt{t+3} + 3})$.
$s'(6) = \lim_{t \to 6} \frac{t + 3 - 9}{(t - 6)(\sqrt{t+3} + 3)}$
$s'(6) = \lim_{t \to 6} \frac{t - 6}{(t - 6)(\sqrt{t+3} + 3)}$.
Take the limit as $t$ goes to 6,
$s'(6) = \lim_{t \to 6} \frac{1}{\sqrt{t+3} + 3} = \frac{1}{?}$
Show more…
Example. An object moving along a straight line has position given by s(t) = √(t+3), where position is measured in meters and time in seconds. Find the (instantaneous) velocity of the object at time t = 6. Explanation. Velocity is the instantaneous rate of change of position with respect to time. We are being asked to find v(6) = s'(6). We have to adjust the definition of the the derivative, namely, replace x with t, s'(6) = limt β†’ 6(s(t) - s(6))/(t - 6) = limt β†’ 6(√(t+3) - 3)/(t - 6). Multiply by (√(t+3) + 3)/(√(t+3) + 3), in order to rationalize the expression, Now simplify the numerator, Combine like-terms, s'(6) = limt β†’ 6((√(t+3) - 3))/(t - 6) ((√(t+3) + 3)/(√(t+3) + 3)). s'(6) = limt β†’ 6(t + 3 - 9)/((t - 6)(√(t+3) + 3)) s'(6) = limt β†’ 6(t - 6)/((t - 6)(√(t+3) + 3)). Take the limit as t goes to 6, s'(6) = limt β†’ 6(1)/(√(t+3) + 3) = (1)/(?)

Added by Robert B.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Example. An object moving along a straight line has position given by s(t)=sqrt(t+3), where position is measured in meters and time in seconds. Find the (instantaneous) velocity of the object at time t=6. Explanation. Velocity is the instantaneous rate of change of position with respect to time. We are being asked to find v(6)=s^(')(6). We have to adjust the definition of the the derivative, namely, replace x with t, s^(')(6)=lim_(t->6)(s(t)-s(6))/(t-6)=lim_(t->6)(sqrt(?-3))/(t-6). Multiply by (sqrt(t+3)+3)/(sqrt(t+3)+3), in order to rationalize the expression, s^(')(6)=lim_(t->6)((sqrt(t+3)-3)/(t-6))((sqrt(t+3)+3)/(sqrt(t+3)+3)). Now simplify the numerator, s^(')(6)=lim_(t->6)(t+3-9)/((t-6)(sqrt(t+3)+3)). Combine like-terms, s^(')(6)=lim_(t->6)((t)/(6))/((t-6)(sqrt(t+3)+3)). Take the limit as t goes to 6 , s^(')(6)=lim_(t->6)(1)/(sqrt(t+3)+3)= Example. An object moving along a straight line has position given by s(t)= /t + 3, where position is measured in meters and time in seconds. Find the (instantaneous) velocity of the object at time t = 6. Explanation. Velocity is the instantaneous rate of change of position with respect to time. We are being asked to find v(6) = s6. We have to adjust the definition of the the derivative,namely,replace x with t, st-s(6 s6=lim -lim t6 97 t6 Multiply by t+3+3 t + 3 3 s'(6)= lim ( t-6 /t + 3 + 3 /t +3 +3 Now simplify the numerator, t+3-9 s6=lim t6 (t=6) (/t+3 +3) Combine like-terms, t s6=lim t6 t6vt+3+3 Take the limit as t goes to 6, 1 s6=lim ++/
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The location P(t) of an object moving in the xy-plane at time t seconds is given by the equations P(t)=(x(t),y(t)), where x(t)=a +9t and y(t)=b +6t, a,b are constants and distances are measured in units of meters. The equations x(t), y(t) describe linear parametrized motion; see section 10.1 of the textbook for review. (a) The location of the object at time t=1 is ( a + 9 , b + 6 ) (b) The average rate of change of x(t) between 1 and 2 seconds is 9 m/s; this is called the average velocity of x(t) on the time interval [1,2]. (c) The instantaneous rate of change of x(t) at time t=1 is 9 m/s; this is called the instantaneous horizontal velocity at time t=1. (d) What is the instantaneous horizontal velocity of the object at time t? 9 (e) The average rate of change of y(t) between 1 and 2 seconds is m/s; this is called the average velocity of y(t) on the time interval [1,2]. (f) The instantaneous rate of change of y(t) at time t=1 is m/s; this is called the instantaneous vertical velocity at time t=1. (g) What is the instantaneous vertical velocity of the object at time t? (h) The line along which the object is moving in the plane has the equation: y= x+ (i) Let d(t) be the distance the object has traveled after t seconds. The formula for d(t) is . (j) The instantaneous rate of change of d(t) at time t is . This is called the speed along the line of motion.

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Transcript

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00:01 All right, guys, so in this problem, i might be reading it wrong.
00:04 It's kind of hard to understand the whole average velocity, the position.
00:09 But so if our position function, s of t, is equal to 2t, sine of 5t, well, if we want to find the velocity function, all we have to do is take the derivative of that.
00:27 So the derivative of this, so we take the derivative of the first of the 2t, so that's 2 times the sign of 5t, plus now we leave 2t alone.
00:40 We take the derivative of 5t, which is the cosine of 5t times the derivative of what we're taking the cosine of, which is 5.
00:52 So i think that's how we're supposed to type it.
00:54 So our velocity function should be 2 times the sign of 5t plus 10 t times the cosine of 5t.
01:13 And that's our velocity function with respect to time.
01:17 All right.
01:17 So now if we want the acceleration function, we just have to take the derivative of our velocity function.
01:24 Well, the derivative of 2 sign of 5t is not negative, is just 2, okay, 2 cosine of 5t times the 5.
01:41 So that's going to be 10 cosine of 5t plus now i got to do the product rule with this.
01:51 So that's going to be 10 times the cosine of 5t plus 10t times.
02:02 Now i'm taking the derivative of cosine, which is negative sign of 5t times 5 because i've got to take the derivative of what's the chain rule with the sign.
02:15 All right.
02:16 So all of this 10 cosine, so this 10 cosine of 5t was this part here.
02:24 Now, adding, i have another 10 cosine of 5t plus, so i have a negative and a 5 and a 10, so it's actually really going to be a negative.
02:47 So it'll be minus 50t times the sign of 5t...
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