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    firstly, just wondering if i've got this right:

    The surface area of a sphere is increasing at the rate of 640 cm^2 s^-1. Find the rate of increase of the radius of the sphere when the radius is 5cm.

    \frac{dA}{dt} = 640

    A = 4\pi r^2 \Rightarrow \frac{dA}{dr} = 8\pi r, \Rightarrow \frac{dr}{dA} = \frac{1}{8\pi r}

    \frac{dr}{dt} = \frac{dr}{dA} \times \frac{dA}{dt} = \frac{1}{8\pi r} = \frac{640}{8\pi r}

    at  r = 5, \frac{dr}{dt} = \frac{640}{40\pi} = 16\pi cms^-1

    2) i) Use Simpson's Rule with 5 ordinates to find the value of

    \displaystyle\int^2_0 ln(1 + sin x) \, dx

    Using x at 0, 0.5, 1.0, 1.5, 2 (4 strips).. i used the rule to find
    yo + 4y1 + 2y2 + 4y3 + y4 = 6.2022... then multiplied this by (0.5/3) = 1.0337 units squared. Anywhere i can check this is right quickly?

    ii) Hence find \displaystyle\int^2_0 ln(1 + sin x)^5 \, dx
    what do i do with answer in i) to get this :p:

    thanks
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    (Original post by Salinger)
    firstly, just wondering if i've got this right:

    The surface area of a sphere is increasing at the rate of 640 cm^2 s^-1. Find the rate of increase of the radius of the sphere when the radius is 5cm.

    \frac{dA}{dt} = 640

    A = 4\pi r^2 \Rightarrow \frac{dA}{dr} = 8\pi r, \Rightarrow \frac{dr}{dA} = \frac{1}{8\pi r}

    \frac{dr}{dt} = \frac{dr}{dA} \times \frac{dA}{dt} = \frac{1}{8\pi r} = \frac{640}{8\pi r}

    at  r = 5, \frac{dr}{dt} = \frac{640}{40\pi} = 16\pi cms^-1

    2) i) Use Simpson's Rule with 5 ordinates to find the value of

    \displaystyle\int^2_0 ln(1 + sin x) \, dx

    Using x at 0, 0.5, 1.0, 1.5, 2 (4 strips).. i used the rule to find
    yo + 4y1 + 2y2 + 4y3 + y4 = 6.2022... then multiplied this by (0.5/3) = 1.0337 units squared. Anywhere i can check this is right quickly?

    ii) Hence find \displaystyle\int^2_0 ln(1 + sin x)^5 \, dx
    what do i do with answer in i) to get this :p:

    thanks
    Question 1) Your \pi should not have jumped to the top!

    Question 2) i) your answer is correct.

    ii) \ln{a^b}=b\ln{a}
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    (Original post by Mr M)
    Question 1) Your \pi should not have jumped to the top!

    Question 2) i) your answer is correct.

    ii) \ln{a^b}=b\ln{a}
    Oops, i did have it as \frac{16}{\pi}cm/s, just did latex wrong :p: is that correct now?

    and ii) will therefore be  5 \times 1.0337?

    thanks
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    (Original post by Salinger)
    Oops, i did have it as \frac{16}{\pi}cm/s, just did latex wrong :p: is that correct now?

    and ii) will therefore be  5 \times 1.0337?

    thanks
    Both sound good to me.
 
 
 
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