Maths Help Calculus

Calculus math help?

ambiguous..

1/3*t^2+t coulombs of charge
guessing this is
C= (1/3)t² + t
although there are many other possibilities.

A = dC/dt = (2/3)t + 1
at t=3
A = (2/3)(3) + 1 = 3 amps

b) 20 = (2/3)t + 1
(2/3)t = 19
t = 19•3/2 = 28.5 sec

Other possibilities
C = (1/(3t²)) + t

C = 1/(3t² + t)

Math help!! (pre-calculus)?

Several ways to do the first one, but perhaps the easiest way is to use the properties of a right triangle. Sketch the graph, which is a line passing through (-3,4) and (1,1). Draw a triangle with Q, M and (1,4) as the end points.

The vertical distance from M to L is 3, and the horizontal distance from Q to L is 4. That means that QML is a 3-4-5 right triangle.

Then build an identical 3-4-5 right triangle from M at the upper left side. Go 4 units to the right along the x axis and then 3 units down. That's the point where P is located.

In the second problem, the center of the circle is at 0,0, so the equation of the circle is x^2 + y^2 = r^2. (If the center were not at the origin, there would be some constants in the equation.)

If x+y=8, the intercepts of that line are (0,8) and (8,0). The midpoint (from the midpoint formula) is at (4,4), and since that's the tangent point, the line from (4,4) to the origin is the radius of your circle. Use the Pythagorean theorem to find the length and plug it into the equation of the circle, which turns out to be x^2 + y^2 = 32.

In the third problem, put both the equations in standard form. You'll see that the only difference is the constant term, which is a function of the radius. So the circles are concentric, both having a center point of (2,0).

Calculus math help please?

At what value(s) of x does f(x) = x^4 – 18x^2 have a critical point where the graph changes from decreasing to increasing?
0 and –3 only
0 and 3 only
–3, and 3 only
0 only

Calculus MATH help?

Do you mean the integral? Get a great maths program called Derive, it does calculus and algebraic stuff. The integral of sin 2x/cos 2x btw is -ln(cos 2x)/2. If you need the method, use the product rule for the integral of sin(2x) * 1/cos(2x). Enjoy yourself!!!

Calculus math help please 4?

I thank you for your time in advance, I will come back and look at your answers.

A piece of molding 184 cm long is to be cut to form a rectangular picture frame. What dimensions will enclose the largest area?



Another question here. http://answers.yahoo.com/question/ask;_ylt=AvzRuhqN.7xErLQ81KWX7t8PzKIX;_ylv=3

Higher maths help. further calculus?

if

y = 1/x^3 - cos2x where x doesnt= 0

Find Dy/Dx


y = x^-3 - cos2x
dy/dx = -3x^-4 - (so this is where i am confused / is the derivative of - cos2x , sin2x or - sin2x ? )


thanks if you understand my question

Calculus Help and Math Summer Packet?

Determine the range and the zeros of: f(x) = 13 - 20x - x2 - 3x4
(The 2 next to the x2 should be an exponent as well as the 4 next to the 3x4)

Find a degree 3 polynomial with zeros -2, 1, and 5 and going through the point (0,-3)

Find the points of intersection of
x2 + y2= 4 and x2 + y2 - 4x - 4y = -4
(the numbers right after the variables are exponents)

Please help if you can. You don't have to answer all three just the ones you can do.

Tough Math, Help please, thank you, Calculus.....?

tan^2(x) + sec(x) - 1 = 0

a) To rewrite this in terms of secant, it is useful to know the identity

tan^2(x) = sec^2(x) - 1

All we have to do is apply this to the tan^2(x) in the equation. This give sus

sec^2(x) - 1 + sec(x) - 1 = 0

We can simplify this, to

sec^2(x) + sec(x) - 2 = 0

b) To find the exact value of sec(x), all we have to do is solve that equation. Note that it is a quadratic in disguise, and if you don't believe me, let y = sec(x). Then we have

y^2 + y - 2 = 0

Factor this, to get

(y + 2) (y - 1) = 0

Therefore, your two equations are y = -2 and y = 1. However, we let y = sec(x), so our two equations are
(i) sec(x) = -2, and (ii) sec(x) = 1

(i) sec(x) = -2, means that if we take the reciprocal of secant, we will get cosine (since cosine is 1/secant). Taking the reciprocal of both sides, we have

cos(x) = -1/2.
Where is cosine equal to -1/2? The answer to that is in quadrants 2 and 3, and x = {7pi/6, 11pi/6}. However, since x lies in quadrant 2, we discard 11pi/6, and x = 7pi/6.

(ii) sec(x) = 1 means
cos(x) = 1, and this is true when
x = 0. However, this does not lie in Quadrant II, so we discard this solution.

Therefore, our only solution is x = 7pi/6.