Results 1 to 9 of 9

Math Help - Related Rates

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    132

    Related Rates

    The altitude of a right-angled triangle is 6 cm and the base is increasing at a constant rate of 2cm/s. At what rate it the hypotenuse increasing when its length is 10 cm?

    No idea how to do this question. But i did approach it. I tried using similar triangles but it didn't work out. Why?

    Let length be y and base be x

    At y=10
    6/y = x/h
    6/10 = x/h
    x = 3/5 h
    dx/dt = 3/5 dh/dt
    thus dh/dt = 10/3 cm/s

    The actual answer is 1.6
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2010
    Posts
    3
    Since it's a right triangle you need to use the pythagorean theorem...

    x^2 + y^2 = z^2

    And since we are dealing with rates you need to derive the equation

    2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}

    The 2's cancel so we're left with

    x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}

    now what we're looking for is how fast the hypotenuse is changing, \frac{dz}{dt}

    and what we are given is
    x = 6

    \frac{dx}{dt} = 0

    y=8 Use pythagorean theorem here to get y

    \frac{dy}{dt} = 2

    and

    z = 10

    then we just solve

    x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}

    0 + 8*2 = 10\frac{dz}{dt}

    \frac{dz}{dt} = \frac{16}{10} = 1.6



    Hope this helps! it's my first time posting a reply so sorry if it's unclear, let me know if you have any questions
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2009
    Posts
    132
    Quote Originally Posted by Zion View Post
    Since it's a right triangle you need to use the pythagorean theorem...

    x^2 + y^2 = z^2

    And since we are dealing with rates you need to derive the equation

    2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}

    The 2's cancel so we're left with

    x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}

    now what we're looking for is how fast the hypotenuse is changing, \frac{dz}{dt}

    and what we are given is
    x = 6

    \frac{dx}{dt} = 0

    y=8 Use pythagorean theorem here to get y

    \frac{dy}{dt} = 2

    and

    z = 10

    then we just solve

    x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}

    0 + 8*2 = 10\frac{dz}{dt}

    \frac{dz}{dt} = \frac{16}{10} = 1.6



    Hope this helps! it's my first time posting a reply so sorry if it's unclear, let me know if you have any questions
    Wow thats great. But i dont really understand this.

    The altitude, the line from the vertex of right angle triangle to its perpendicular opposite is 6. And it is given dx/dt = 2 m/s

    Please do help me again!
    Attached Thumbnails Attached Thumbnails Related Rates-diagram.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    Quote Originally Posted by Lukybear View Post
    Wow thats great. But i dont really understand this.

    The altitude, the line from the vertex of right angle triangle to its perpendicular opposite is 6. And it is given dx/dt = 2 m/s

    Please do help me again!
    Here's an easier wayI think.
    First of all redraw your diagram. Draw it (I don't know how to here so I'm giving you a verbal description - sorry) with the right angle in the bottom left corner. The height (vertical line on left) is 6cm, call the base x and the hypotenuse h. Your initial diagram was incorrect.
    Tell me when you are done and I'll help with the next step.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    You still there? Happy to walk you through the steps of related rates problems if interested.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jul 2009
    Posts
    132
    Quote Originally Posted by Debsta View Post
    Here's an easier wayI think.
    First of all redraw your diagram. Draw it (I don't know how to here so I'm giving you a verbal description - sorry) with the right angle in the bottom left corner. The height (vertical line on left) is 6cm, call the base x and the hypotenuse h. Your initial diagram was incorrect.
    Tell me when you are done and I'll help with the next step.
    But if just going on my diagram how would you do this problem.

    As I am pretty sure that the altitude is that line joining the vertex to perpendicular opposite.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    Quote Originally Posted by Lukybear View Post
    But if just going on my diagram how would you do this problem.

    As I am pretty sure that the altitude is that line joining the vertex to perpendicular opposite.
    A triangle actually has 3 altitudes, depending on which side you define to be the base - altitude and "base" are perpendicular. Your diagram is incorrect for this problem.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jul 2009
    Posts
    132
    Quote Originally Posted by Debsta View Post
    A triangle actually has 3 altitudes, depending on which side you define to be the base - altitude and "base" are perpendicular. Your diagram is incorrect for this problem.
    So your saying that the hypotenuse is between the base and the altitude?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    Quote Originally Posted by Lukybear View Post
    So your saying that the hypotenuse is between the base and the altitude?
    yes
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Related Rates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 26th 2009, 08:54 PM
  2. Rates and Related Rates!!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 2nd 2008, 10:53 AM
  3. Another Related Rates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 13th 2008, 06:32 AM
  4. Related Rates
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 12th 2008, 05:49 PM
  5. rates and related rates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 29th 2007, 09:51 PM

Search Tags


/mathhelpforum @mathhelpforum