How do you do part a) ?
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Originally Posted by princessmath How do you do part a) ? Hi there, You have x^2 + y^2 = 200 differentiating them implicitly gives you (x^2)' + (y^2)' = (200)' which gives you 2x(dx/dt) + 2y = 0 ... now, plug in the values and solve for dx/dt, hope this help, dokrbb
Last edited by dokrbb; April 15th 2013 at 05:32 AM.
Originally Posted by dokrbb Hi there, You have x^2 + y^2 = 200 differentiating them implicitly gives you (x^2)' + (y^2)' = (200)' which gives you 2x(dx/dt) + 2y = 0 ... now, plug in the values and solve for dx/dt, hope this help, dokrbb This is incorrect. x and y are both functions of t, so when you differentiate both sides, you should get NOW you can substitute in the known values and solve for .
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