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Math Help - tangent to the curve

  1. #1
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    tangent to the curve

    i think i'm making a silly mistake somewhere, because i should be able to do this!

    A curve has the equation y= 3x+11 (all square rooted)
    The point P on the curve has x-coordinate 3

    Show that the tangent to the curve at P has the equation 3x-4root5y + 31=0

    any help would be greatly appreciated, i cant seem to see where ive gone wrong!
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  2. #2
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    Quote Originally Posted by Oranges&Lemons View Post
    i think i'm making a silly mistake somewhere, because i should be able to do this!

    A curve has the equation y= 3x+11 (all square rooted)
    The point P on the curve has x-coordinate 3

    Show that the tangent to the curve at P has the equation 3x-4root5y + 31=0

    any help would be greatly appreciated, i cant seem to see where ive gone wrong!
    Since you know you've gone wrong, you're obviously getting an answer different to the given one. It would be useful to see your working, that is, to see how you got your answer. Did you follow this basic approach:

    1. \frac{dy}{dx} = \frac{3}{2 \sqrt{3x + 11}} therefore the gradient of the tangent to the curve at P is given by m = .....

    2. A point on the tangent to the curve at P is (3, 20).

    3. The equation of any line (including a tangent) has the general form y - y_1 = m(x - x_1).

    4. Conclude that the given answer of 3x - 4 \sqrt{5} y + 31 = 0 is wrong - that the 31 should be 80 \sqrt{5} - 9 ....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Since you know you've gone wrong, you're obviously getting an answer different to the given one. It would be useful to see your working, that is, to see how you got your answer. Did you follow this basic approach:

    1. \frac{dy}{dx} = \frac{3}{2 \sqrt{3x + 11}} therefore the gradient of the tangent to the curve at P is given by m = .....

    2. A point on the tangent to the curve at P is (3, 20).

    3. The equation of any line (including a tangent) has the general form y - y_1 = m(x - x_1).

    4. Conclude that the given answer of 3x - 4 \sqrt{5} y + 31 = 0 is wrong - that the 31 should be 80 \sqrt{5} - 9 ....

    ok...its the differentiating ive done wrong! i differentiated it as...

    0.5(3x+11)to the power -0.5 then times 3....


    so 1.5(3x+11)to the power -0.5
    ive used the wrong rule havent i...
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    hang on are you saying the answer given on the paper is wrong? its not -31 on the end?
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    Quote Originally Posted by Oranges&Lemons View Post
    ok...its the differentiating ive done wrong! i differentiated it as...

    0.5(3x+11)to the power -0.5 then times 3....


    so 1.5(3x+11)to the power -0.5
    ive used the wrong rule havent i...
    Your derivative is correct. 1.5 is the same as 3/2. Raising to -0.5 is the same as taking the reciprocal square root.
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  6. #6
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    Quote Originally Posted by Oranges&Lemons View Post
    hang on are you saying the answer given on the paper is wrong? its not -31 on the end?
    Wait a moment. Let me just go back and read exactly what I wrote ......

    Quote Originally Posted by mr fantastic View Post
    [snip]
    4. Conclude that the given answer of 3x - 4 \sqrt{5} y + 31 = 0 is wrong - that the 31 should be 80 \sqrt{5} - 9 ....
    Quote Originally Posted by Oranges&Lemons View Post
    hang on are you saying the answer given on the paper is wrong? its not -31 on the end?
    Yep, that's what I said alright.
    WAIT WAIT WAIT. WHAT I SAID IS WRONG. LET ME GO BACK AND EDIT MY WORK.
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  7. #7
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    it's just clicked! thanks.
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  8. #8
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    This is what I should have written:

    Quote Originally Posted by mr fantastic on an alternative world-line
    Since you know you've gone wrong, you're obviously getting an answer different to the given one. It would be useful to see your working, that is, to see how you got your answer. Did you follow this basic approach:

    1. \frac{dy}{dx} = \frac{3}{2 \sqrt{3x + 11}} therefore the gradient of the tangent to the curve at P is given by m = .....

    2. A point on the tangent to the curve at P is (3, \sqrt{20}), that is, (3, 2 \sqrt{5}).

    3. The equation of any line (including a tangent) has the general form y - y_1 = m(x - x_1).

    4. Conclude that the given answer of 3x - 4 \sqrt{5} y + 31 = 0 is correct.
    When I (the Mr F on this world-line) was getting the y-coordinate of P in 2., I looked at the function you posted but forgot to read the (all square rooted) bit at the end .... In my defence, an easy mistake of misreading to make ......
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  9. #9
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    Quote Originally Posted by Oranges&Lemons View Post
    it's just clicked! thanks.
    So all's well that ends well.
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  10. #10
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    it is +31 though because the y coordinate at P is root 20 not just 20 thanks.
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  11. #11
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    Quote Originally Posted by Oranges&Lemons View Post
    it is +31 though because the y coordinate at P is root 20 not just 20 thanks.
    Yep, see post #8. So, where was your mistake?
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    Yep, see post #8. So, where was your mistake?
    oh yes, my bad! didn't see that post sorry. my mistake was some sort of calculator error when i subbed in the 3...it didnt click that 3/2(3x+11) to the power -1/2 is the same as 3/2root3x+11 either.
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  13. #13
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    could you help me with this too by any chance?

    i need to differentiate this to do a question...

    x^(5/2)ln(x/4)
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  14. #14
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    Quote Originally Posted by Oranges&Lemons View Post
    could you help me with this too by any chance?

    i need to differentiate this to do a question...

    x^(5/2)ln(x/4)
    Use the product rule: f = zy, f' = z'y + y'z

    Let z=x^{(5/2)} and y=ln(x/4).
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  15. #15
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    Quote Originally Posted by colby2152 View Post
    Use the product rule: f = zy, f' = z'y + y'z

    Let z=x^{(5/2)} and y=ln(x/4).
    of courseeeee. thanks!
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