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Math Help - Looking for a little help plz

  1. #1
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    Looking for a little help plz

    I am in a calculus class and am stuck on two problems that have something to do with derivatives. I think im on the right trak when starting them but tend to fall of when getting to the expanded portions of the problem. Maybe im making them harder than they really are, but here are the two problems:

    1.) Car A is traveling at 50 mi/h and car B is traveling north at 60 mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?


    2.) The twice-differentiable function f is defined for all real numbers and satisfies the following conditions: f(0) = 2, f'(0) = -4, and f"(0) = 3.

    A.) The function g is given by g(x) = e^ax + f(x) for all real numbers, where a is a constant. Find g'(0) and g"(0) in terms of a.

    B.) The function h is given by h(x) = cos(kx)*f(x) for all real numbers, where k is a constant. Find h'(x) and write an equation for the line tangent to the graph of h at x = 0

    Problem #2 I think I have right, but being able to compare with someone elses work would be good, that way I can see what I may have did wrong or could have done differently. Thanks for any help you all can give

    ' = Derivative.
    " = Double Darivative.
    *= Multiplication
    .
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  2. #2
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    Quote Originally Posted by Munroe View Post
    I am in a calculus class and am stuck on two problems ...:
    ...

    2.) The twice-differentiable function [FONT=Fixedsys][FONT=Arial Black]f [FONT=Verdana]is defined for all real numbers and satisfies the following conditions: f(0) = 2, f'(0) = -4, and f"(0) = 3.

    A.) The function g is given by g(x) = e^ax + f(x) for all real numbers, where a is a constant. Find g'(0) and g"(0) in terms of a.

    B.) The function h is given by h(x) = cos(kx)*f(x) for all real numbers, where k is a constant. Find h'(x) and write an equation for the line tangent to the graph of h at x = 0

    Problem #2 I think I have right, but being able to compare with someone elses work would be good, ...
    Hello,

    I'll give only the results of problem #2 so you can compare your answers with what I've got:

    to A):

    g(x) = e^(ax) + f(x)

    g'(x) = a*e^(ax) + f'(x) ==> g'(0) = a - 4

    g''(x) = a*e^(ax) + f''(x) ==> g''(0) = a + 3

    to B):

    h(0) = cos(k*0)*f(0) = 1 * 2 = 2

    h'(x) = cos(k*x)*f'(x) - f(x) * sin(k*x) * k ==> h'(0) = (-2) * 0 + cos(0)*(-4) = -4

    Therefore the equation of the tangent is:

    t: y = -4x + 2

    EB
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  3. #3
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    Quote Originally Posted by Munroe View Post
    I am in a calculus class and am stuck on two problems that have something to do with derivatives. ...

    1.) Car A is traveling at 50 mi/h and car B is traveling north at 60 mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?
    ...
    Hello,

    I'l add a few more facts so that I can do this problem:

    i) The roads are perpendicular to each other.
    ii) Car A is going east.

    I use a coordinate system with the intersection as origin. The car A has the position (-0.3, 0) at the time t = 0 and the car B has the position (0, -0.4) at t = 0.

    Car A travels in t hours a distance of w(t) = 50 * t (in miles)
    Car B travels in t hours a distance of w(t) = 60 * t (in miles)

    The distance of car A from the intersection is: -0.3 + 50*t
    The distance of car B from the intersection is: -0.4 + 60*t

    Now you can calculate the distance between the two cars (use the Pythagoran theorem):

    d = (-0.3 + 50*t) + (-0.4 + 60*t) = 6100t - 78t + 0.25
    Therefore

    d(t) = √(6100t - 78t + 0.25)

    The rate of approaching is d'(t)

    d'(t)=(1/2)*(6100t - 78t + 0.25)^(-1/2) * (12200t - 78) = (6100t - 39)/(√(6100t - 78t + 0.25))

    The approaching stops if d'(t) = 0 ==> t = 39/6100 hours t ≈ 23.02 seconds.

    There is a distance between the two cars of 0.0256.. miles, approximately 135.2 feet.

    EB
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  4. #4
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    Thnx for the help EB. If anyone else could elaborate on problem number 2 for me also so I can try to match my answers. EB did it well, I just wanna see how someone else breaks it down so I can see another view of it. Seeing the problem done more than once helps me out alot
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  5. #5
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    Quote Originally Posted by Munroe View Post
    ...
    2.) The twice-differentiable function [FONT=Fixedsys][FONT=Arial Black]f [FONT=Verdana]is defined for all real numbers and satisfies the following conditions: f(0) = 2, f'(0) = -4, and f"(0) = 3.

    A.) The function g is given by g(x) = e^ax + f(x) for all real numbers, where a is a constant. Find g'(0) and g"(0) in terms of a.

    B.) The function h is given by h(x) = cos(kx)*f(x) for all real numbers, where k is a constant. Find h'(x) and write an equation for the line tangent to the graph of h at x = 0
    ...
    Hi,

    I have attached an image to show you what I've done with #2, A
    Attached Thumbnails Attached Thumbnails Looking for a little help plz-munroe_a.gif  
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  6. #6
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    Quote Originally Posted by Munroe View Post
    ...


    2.) The twice-differentiable function [FONT=Fixedsys][FONT=Arial Black]f [FONT=Verdana]is defined for all real numbers and satisfies the following conditions: f(0) = 2, f'(0) = -4, and f"(0) = 3.

    A.) The function g is given by g(x) = e^ax + f(x) for all real numbers, where a is a constant. Find g'(0) and g"(0) in terms of a.

    B.) The function h is given by h(x) = cos(kx)*f(x) for all real numbers, where k is a constant. Find h'(x) and write an equation for the line tangent to the graph of h at x = 0
    ...
    Hi,

    with #2, B you have to use the same values for f(0), f'(0) and f''(0) as in #2, A.
    Attached Thumbnails Attached Thumbnails Looking for a little help plz-munroe_b.gif  
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