Kevin Said:

I need help with a few ap calculus problems.?

We Answered:

/
1)v=6t-2
v(2)=10
a=6 at all time

Alvin Said:

Please help with AP Calculus AP Problems?

We Answered:

Visit these sites to learn more about mathematical formulas and concepts that can help you in regards to your problem:

??http://www.math.com/tables/index.html
??http://www.hsd.k12.or.us/glencoe/staff/a…
??http://www.regentsprep.org/Regents/Math/…

Jo Said:

few more ap calculus problems?

We Answered:

f(x) = 4 - x^2

Slope of f(x) is f '(x) = -2x.
At the point (p, 4 - p^2), the slope is -2p.

Let tangent to graph be : y = mx + b
Substituting (x, y) = (p, 4 - p^2) and m = -2p gives :
4 - p^2 = -2p(p) + b
Therefore, b = 4 + p^2
Thus, tangent line is : y = -2px + 4 + p^2

Tangent line touches X-axis when y = 0.
If y = 0, then x = (4 + p^2) / (2p), which is the base of the triangle.

Tangent line touches Y-axis when x = 0.
If x = 0, then y = 4 + p^2, which is the height of the triangle.

Area, A(p) = (1/2)*base*height
= (1/2)*[(4 + p^2) / (2p)]*(4 + p^2)
= (4 + p^2)^2 / (4p)

(a) A(2) = (4 + 2^2)^2 / (4*2) = 8

(b) A(p) = (4 + p^2)^2 / (4p)
Take the derivative and set it equal to zero.
dA(p)/dp = [4p(2)(4 + p^2)(2p) - 4(4 + p^2)^2] / (16p^2) = 0
Solving gives : p = 2/?(3), which is the value of p when A(p) is a minimum.
This minimum area = 32?(3)/9.

Arthur Said:

3 AP Calculus problems, need urgent help!?

We Answered:

1

To find a(t), you need to find first derivative of v(t)

v(t) = 7 - (1.01)^(-t²)

a(t) = v'(t) = 0 - (1.01)^(-t²)*(-2t)*ln(1.01)
a(t) = 2t ln(1.01) * (1.01)^(-t²)

a(3) = 6 ln(1.01) * (1.01)^(-9)
a(3) = 0.054587903

Answer: b) 0.055

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#2

x(3) = x(0) + ??³ (1+t²)^? dt

Now this function cannot really be integrated by hand. You need to use a graphing/programmable calculator to find the definite integral. This gives us:

x(3) = 2 + 4.51153246 = 6.51153246

Answer: d) 6.512

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#3.

f(3) = 15, f(6) = 3

Since g(x) = f?¹(x), then
g(15) = 3, g(3) = 6

and
g'(3) = 1/f'(6) = 1/-2 = -1/2

Answer: a) -1/2

Alvin Said:

Can anyone help me with these two AP Calculus problems?

We Answered:

comments : 1(a) let u = x² + 1 for the integration...2(2) this is obviously in error ...at what x value is the derivative = to 6 and the 2nd derivative = to 18?.....from 1) you know the form of f...[a/3] x^3 + [b/2] x^2 + c...2) should find a & b values and 3) will let you find the value of c.