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Math Help - Need help with these 4 questions...

  1. #1
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    Need help with these 4 questions...

    1. I am supposed to find the derivative of (sqrt) 3x using the (lim h ->0) f'(x) = (f(x+h) - f(x))/h equation and here is what i got:

    ((sqrt) 3(x+h) - (sqrt) 3x) / h
    then i got ((sqrt) 3x + 3h) - (sqrt) 3x) / h
    I then used the conjugate of this and got:
    (3x + 3h - 3x) / h((sqrt) 3x + 3h + (sqrt) 3x)

    the 3x in the numerator cancels and the h in 3h is factored out. so i then get:
    3 / ((sqrt) 3x + 3h) + (sqrt) 3x.
    when h -> 0, i got

    3 / 2(sqrt) 3x.

    I know I am making a dumb mistake somewhere and know that the answer is suppose to be: (3/2) x^(-1/2)

    2. This is a problem involving an airplane taking off...Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10/9)t^2 where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel in that time?

    The problem I have with this is I do not know where to start nor end.

    3. A particle moves on a coordinate line such that it has a position
    s = 2t^3 - 4t^2 + 7 on the interval 0 =< t =< 3, with s measured in meters and t measured in seconds. Find the particle's velocity and acceleration at the endpoints of the interval. This is what i got.

    s'(t) = 6t^2 - 8t
    s''(t) = 12t - 8
    v(0) = 6(0)^2 - 8(0) = 0 m/sec.
    v(3) = 54 - 24 = 30 m/sec
    A(0) = 12(0) - 8 = -8 m/sec
    A(3) = 12(3) - 8 = 28 m/sec

    I just need to know if this is right.

    4. I am suppose to find the derivative of 2t^(3/2) + 3e^2

    e is the natural log.

    I know the first part is 3t^(1/2), but i do not know about the second part. The answer in the book said just 3t^(1/2). But I thought that I am suppose to get something for an answer for the second part and this is where i got confused.

    Thank you in advance.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by driver327 View Post
    1. I am supposed to find the derivative of (sqrt) 3x using the (lim h ->0) f'(x) = (f(x+h) - f(x))/h equation and here is what i got:

    ((sqrt) 3(x+h) - (sqrt) 3x) / h
    then i got ((sqrt) 3x + 3h) - (sqrt) 3x) / h
    I then used the conjugate of this and got:
    (3x + 3h - 3x) / h((sqrt) 3x + 3h + (sqrt) 3x)

    the 3x in the numerator cancels and the h in 3h is factored out. so i then get:
    3 / ((sqrt) 3x + 3h) + (sqrt) 3x.
    when h -> 0, i got

    3 / 2(sqrt) 3x.

    I know I am making a dumb mistake somewhere and know that the answer is suppose to be: (3/2) x^(-1/2)
    The derivative of \sqrt{3x} is \frac{3}{2\sqrt{3x}}=\frac{\sqrt{3}}{2}x^{-1/2}


    2. This is a problem involving an airplane taking off...Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10/9)t^2 where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel in that time?

    The problem I have with this is I do not know where to start nor end.
    Since D=\frac{10}9t^2, we're looking for t when D^{\prime}=\frac{500}{9} (this is because 200 km/hr = 500/9 m/s). Then you can find the distance traveled in that time.

    3. A particle moves on a coordinate line such that it has a position
    s = 2t^3 - 4t^2 + 7 on the interval 0 =< t =< 3, with s measured in meters and t measured in seconds. Find the particle's velocity and acceleration at the endpoints of the interval. This is what i got.

    s'(t) = 6t^2 - 8t
    s''(t) = 12t - 8
    v(0) = 6(0)^2 - 8(0) = 0 m/sec.
    v(3) = 54 - 24 = 30 m/sec
    A(0) = 12(0) - 8 = -8 m/sec^2
    A(3) = 12(3) - 8 = 28 m/sec^2

    I just need to know if this is right.
    Note my correction in red. Otherwise, everything else is correct.

    4. I am suppose to find the derivative of 2t^(3/2) + 3e^2

    e is the natural log.

    I know the first part is 3t^(1/2), but i do not know about the second part. The answer in the book said just 3t^(1/2). But I thought that I am suppose to get something for an answer for the second part and this is where i got confused.

    Thank you in advance.
    3e^2 is a constant, where e=2.7182818.... Thus the derivative of 3e^2 is zero.

    So the answer in the book is correct.

    Does this make sense?
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by driver327 View Post
    1. I am supposed to find the derivative of (sqrt) 3x using the (lim h ->0) f'(x) = (f(x+h) - f(x))/h equation and here is what i got:

    ((sqrt) 3(x+h) - (sqrt) 3x) / h
    then i got ((sqrt) 3x + 3h) - (sqrt) 3x) / h
    I then used the conjugate of this and got:
    (3x + 3h - 3x) / h((sqrt) 3x + 3h + (sqrt) 3x)

    the 3x in the numerator cancels and the h in 3h is factored out. so i then get:
    3 / ((sqrt) 3x + 3h) + (sqrt) 3x.
    when h -> 0, i got

    3 / 2(sqrt) 3x.

    I know I am making a dumb mistake somewhere and know that the answer is suppose to be: (3/2) x^(-1/2)
    You get a correct answer

    f\left( x \right) = \sqrt {3x}

    f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {3\left( {x + h} \right)}  - \sqrt {3x} }}{h} =

    = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {3\left( {x + h} \right)}  - \sqrt {3x} } \right)\left( {\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} } \right)}}{{h\left( {\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} } \right)}} =

    = \mathop {\lim }\limits_{h \to 0} \frac{{3\left( {x + h} \right) - 3x}}<br />
{{h\left( {\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{3}{{\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} }} =

    = \frac{3}{{\sqrt {3\left( {x + 0} \right)}  + \sqrt {3x} }} = \frac{3}<br />
{{2\sqrt {3x} }}.

    Or \frac{3}{{2\sqrt {3x} }} = \frac{{3\sqrt 3 }}{{2\sqrt 3 \sqrt x \sqrt 3 }} = \frac{{3\sqrt 3 }}{{2 \cdot 3\sqrt x }} = \frac{{\sqrt 3 }}{{2\sqrt x }}.
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  4. #4
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    Quote Originally Posted by DeMath View Post
    You get a correct answer

    f\left( x \right) = \sqrt {3x}

    f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {3\left( {x + h} \right)}  - \sqrt {3x} }}{h} =

    = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {3\left( {x + h} \right)}  - \sqrt {3x} } \right)\left( {\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} } \right)}}{{h\left( {\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} } \right)}} =

    = \mathop {\lim }\limits_{h \to 0} \frac{{3\left( {x + h} \right) - 3x}}<br />
{{h\left( {\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{3}{{\sqrt {3\left( {x + h} \right)}  + \sqrt {3x} }} =

    = \frac{3}{{\sqrt {3\left( {x + 0} \right)}  + \sqrt {3x} }} = \frac{3}<br />
{{2\sqrt {3x} }}.

    Or \frac{3}{{2\sqrt {3x} }} = \frac{{3\sqrt 3 }}{{2\sqrt 3 \sqrt x \sqrt 3 }} = \frac{{3\sqrt 3 }}{{2 \cdot 3\sqrt x }} = \frac{{\sqrt 3 }}{{2\sqrt x }}.
    With 1., Now I realized my mistake. I gotten a little confused about the 3.

    With 4., I had momentarily forgot the natural log was a constant instead of a variable.

    With 2. I am still a little confused. The way I attempted this was I took the derivative of (10/9))(t^2)and got (20/9)(t). I tried time conversion to solve (1 hour = 3600 secs.). My real difficulty is putting all of this together. please help.

    Besides 2. I now understand the errors I have made.
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