Recent content by parabol

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    Circular motion of rope and ball

    Homework Statement A motor is used to rotate a ball attached to a rope in a vertical plane. The mass of the ball is 2kg and the length of the rope is 3m. Ignoroing air resistance and the mass of the rope, calculate: a) The minimum motor speed in rpm that will maintain the ball in a...
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    Finding the Gradient of ln(tan2x+secx) Using Substitution | Homework Help

    I'm still not 100% with this question. I have been re-working on it and have now come up with this answer. \frac{dy}{dx}=\frac{\frac{2(sin2x)^2}{(cos2x)^2}+\frac{sinx}{(cosx)^2}+2}{\frac{sin2x}{cos2x}+\frac{1}{cosx}} Does this look right?
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    Finding the Gradient of ln(tan2x+secx) Using Substitution | Homework Help

    Thanks for the input. Have I started off using the method by using the chain rule? Do I just leave it as: (\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1} ?
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    Couple of simple (I hope) questions

    Thanks for your help. I reckon I have it sorted now. Cheers
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    Finding the Gradient of ln(tan2x+secx) Using Substitution | Homework Help

    Homework Statement By using cos and sin subs for tan and sec, find the gradient of: ln(tan2x+secx) Homework Equations tanx=sinx/cosx secx=1/cosx The Attempt at a Solution Substituting y=ln(\frac{sin2x}{cos2x}+\frac{1}{cosx}) Using the chain rule let: z=...
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    Couple of simple (I hope) questions

    Thanks for the reply. Happy with the first part. But for the 2nd. Are you saying that I have carried it out incorrectly by findging the first and second derivative and then integrating them and working out thorugh the boundary values?
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    Couple of simple (I hope) questions

    Homework Statement 1st problem I'm struggling to answer this question using my calculus workbook. Find the point of intersection of the lines: y_{1}=3x+2 3y_{2}=16-x I can solve this question as a simultaneous equation, but, as it is in my calculus question paper believe it to...
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    Find the gradient of y with respect to x

    Things are starting to become a bit clearer, I think/hope. In this question then, is K=3\sqrt{\theta^{2}+1}? And if it is then \frac{dy}{dx}=(3\sqrt{\theta^{2}+1})\times(-x.sin(x^{2}+2\theta)) If it is I can see why you siad I had complicated it. I just wouldn't know how to make it less so.
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    Find the gradient of y with respect to x

    that would be secxtanx. I do not understand the above statement. Could you elaborate? Thanks
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    Find the gradient of y with respect to x

    Thanks for your help tiny-tim. But you have completely thrown me there. The texts I am working off are, well, limited and teach the basics. They essentially cover, sum, product, chain rule and quotient rule and then some higher derivatives, so I'm not certain what techniques I should be using...
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    Find the gradient of y with respect to x

    Thanks tiny-tim A touch of progress then. Utilsing the chain rule I have differentiated the denominator. let v = x2 + 2\theta, let u = cos v let y = 1/2u \frac{dv}{dx}= 2x ... \frac{du}{dv}= -sin.v = -sin(x^{2}+2\theta) ... \frac{dy}{du}= \frac{1}{2}...
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    Find the gradient of y with respect to x

    Homework Statement Find the gradient of y with respect to x: y=\frac{3\sqrt{\theta^{2}+1}}{\frac{1}{2}cos(x^{2}+2\theta)} Well, I am at a complete loss where to start with this. The learning package I have in all its examples and text has no similar worded examples or utilises 2...
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    Sanity Check: Solving for t in x=12(1-e^-t/RC) with Inversion Method

    Thanks Mark. Its just that I tried the invert first off and got completely different colutions once I had substitued values in (why I got confused) .
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    Sanity Check: Solving for t in x=12(1-e^-t/RC) with Inversion Method

    Homework Statement Transpose for t, x=12(1-e^(-t/RC)) I can't get it out of my head but, the soltuion I have come up with doesn't seem right. Solution is, t=-RC ln(1-x/12) Thanks in advance, Realted, but not as important as the sanity check. For my own mind, could someone explain to me...
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    Solving for x: "1/2*log(5)(x^2-1)=1/4+1/2log(5)(x-1)

    All sorted now. Thanks very much for the sanity checks with this.
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