# Differentiation

• Sep 24th 2008, 12:09 AM
andrew2322
Differentiation
Hello all, i'm new to differentiation and i'm totally confused.

1.) A space vehicle moves so that the distance travelled over its first minutes of motion is given by y = 4t^4, where y is the distance travelled in metres and t the time in seconds. By finding the gradient of the chord between points where t = 4 and t = 5, estimate the speed of the space vehicle when t = 5.

2.) Could someone help me define the "limits" and "derivatives" in this topic

Thanks.
• Sep 24th 2008, 02:02 AM
Prove It
Quote:

Originally Posted by andrew2322
Hello all, i'm new to differentiation and i'm totally confused.

1.) A space vehicle moves so that the distance travelled over its first minutes of motion is given by y = 4t^4, where y is the distance travelled in metres and t the time in seconds. By finding the gradient of the chord between points where t = 4 and t = 5, estimate the speed of the space vehicle when t = 5.

2.) Could someone help me define the "limits" and "derivatives" in this topic

Thanks.

A derivative is the rate of change of a function, and is represented by the slope of the graph.

In question 1), the rate of change of distance with respect to time is SPEED.

To estimate the speed at t = 5, remember that the gradient/slope of a line is given by

$\displaystyle \frac{y_2 - y_1}{x_2 - x_1}$.

At $\displaystyle t=4$, $\displaystyle y=4\times 4^4 = 1024$.

At $\displaystyle t=5$, $\displaystyle y=4 \times 5^4 = 2500$.

So $\displaystyle x_1=4$, $\displaystyle x_2 = 5$, $\displaystyle y_1 = 1024$, $\displaystyle y_2 = 2500$.

So the slope (or estimate of the speed) is given by

$\displaystyle \frac{2500 - 1024}{5 - 4} = \frac{1476}{1} = 1476$.

For question 2), like I said, SPEED is the derivative of this function.

Supposing the distance between the two values of t was smaller. If this change in t (denoted by $\displaystyle \Delta t$) was closer and closer to 0, the approximation of the slope gets better and better.

So the limit is the limit as $\displaystyle \Delta t$ tends to 0 of the gradient formula.
• Sep 24th 2008, 02:51 AM
andrew2322
thanks
thanks for the reply, but the back of my book says 2000 m/s, i got your answer as well, so i'm a little confused.
• Sep 24th 2008, 02:56 AM
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Quote:

Originally Posted by andrew2322
thanks for the reply, but the back of my book says 2000 m/s, i got your answer as well, so i'm a little confused.

Don't take BOB's word as law... there are often mistakes in printing.

Can you see anything wrong with your logic? Or mine?

Did you copy down the question properly?
• Sep 24th 2008, 02:57 AM
andrew2322
Hey
Yeah, i'm a little confused because they also say at t = 5, they don't specifically mention anything to do with the approximation of t = 4 and t = 5 seconds... do they?
• Sep 24th 2008, 02:58 AM
hockey777
I don't think he got an answer for the second part. The first part was just a simple slope and demonstration of what the problem was looking for. The second part asks for the derivative at t=5. Try that out.
• Sep 24th 2008, 03:07 AM
Prove It
Quote:

Originally Posted by andrew2322
Yeah, i'm a little confused because they also say at t = 5, they don't specifically mention anything to do with the approximation of t = 4 and t = 5 seconds... do they?

OK the question is wrong...

In order to get 2000, you would need to find the derivative of this function at $\displaystyle t=5$, not approximate it using those two points.

Remember that a derivative is the limit of the gradient formula?

$\displaystyle y(t) = 4t^4$,

$\displaystyle y(t+\Delta t) = 4(t + \Delta t)^4 = 4(t^4 + 4t^3 \Delta t + 6t^2 \Delta t ^2 + 4t \Delta t^3 + \Delta t^4)$
$\displaystyle = 4t^4 + 16t^3 \Delta t + 24t^2 \Delta t ^2 + 16t \Delta t^3 + 4\Delta t^4$

So $\displaystyle \frac{dy}{dt} = \lim_{\Delta t \to 0} \frac{y(t+\Delta t) - y(t)}{\Delta t}$

$\displaystyle = \lim_{\Delta t \to 0} \frac{16t^3 \Delta t + 24t^2 \Delta t ^2 + 16t \Delta t^3 + 4\Delta t^4}{\Delta t}$

$\displaystyle = \lim_{\Delta t \to 0} 16t^3 + 24t^2 \Delta t + 16t \Delta t^2 + 4\Delta t^3$

$\displaystyle = 16t^3$

Evaluating the derivative at $\displaystyle t=5$ gives

$\displaystyle 16\times 5^3 = 2000$.