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Math Help - Differentiation

  1. #1
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    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.
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    Quote Originally Posted by andrew2322 View Post
    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

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

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

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

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

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

    \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 \Delta t) was closer and closer to 0, the approximation of the slope gets better and better.

    So the limit is the limit as \Delta t tends to 0 of the gradient formula.
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    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.
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    Quote Originally Posted by andrew2322 View Post
    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?
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    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?
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    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.
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    Quote Originally Posted by andrew2322 View Post
    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 t=5, not approximate it using those two points.

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

    y(t) = 4t^4,

    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)
     = 4t^4 + 16t^3 \Delta t + 24t^2 \Delta t ^2 + 16t \Delta t^3 + 4\Delta t^4

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

     = \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}

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

     = 16t^3

    Evaluating the derivative at t=5 gives

     16\times 5^3 = 2000.
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