Recent content by mathmann

Thank you very much Kuno.. great help.

The answer says d^2y/dx^2 = -20x^3/y^9. ??

I understand now.. thanks for the help.

s(t) = t^2 - 2/t + 1, is the object speeing up at 4s? v(t) = 1.04, a(t) the numerator ended up as a 0. Perhaps I made a calculating error but I went over it a couple times.

Just wondering how you take the second derivative when using the quotient rule. After using the quotient rule to get my first derivative, I tried again and the numerator ended up as 0.

Homework Statement For the Relation defined by x^5 + y^5 = 5 show that d2y/dx2. Homework Equations The Attempt at a Solution x^5 + y^5 = 5 5x^4 + 5^4dy/dx = 0 d2y/dx2 = - 20x^3/20y^3 ??

Thanks Halls of Ivy For volume of the water I got 30 000 cm(cubed) and for the rate I got 0.3 cm(cubed)/s when the height is 10 cm. Is this now correct? FYI : It is a intro to calculus class.

How do I make it a function?

Homework Statement A triangular prism has end peices in the shape of inverted isosceles traingles with bases 60cm and heights 40 cm . It is 4m long and water is being pumped into it at a rate of 9 L/s. How fast is the level of water rising when the water is 10cm deep. Homework Equations...

thanks for the help. Just wanted to verify one thing though, don't you muliply -1 by the (x + 2) making them both negative. y - 3 = -1/6 (x + 2) 6y - 18 = -x -2 x + 6y -16 = 0

I did, should be -1/6 typo above. sorry Is it correct now?

its not the reciprocal? my current answer in standard form is.. x + 6y - 16 = 0

Is it the reciprocal? 1 _ ? 6

Homework Statement Find the equation of the normal line to the curve y(sqrd) - x(sqrd)y + 3 = 0 at the point (-2,3). (standard form) Homework Equations y - y = m(x - x) The Attempt at a Solution dy __ = -6 dx y - 3 = -6(x + 2) 6x + y +9 = 0

What is the difference between a normal line and a tangent line?