easy calculus problems

[匿名]

2010-03-05 3:37 pm

1.) Find a formula for the derivative dy/dx given that x^2 + 4xy+y^2=13. It is acceptable to leave your answer in terms of both x and y.

2.) determine a formula for the line which is tangent to the graph of y=arctan(e^6x) at its y-intercept.

3.)Find a formula for f '(x) given that f(x)=ln(x^5 e^x3(x^4+3)^6).

4.) Suppose that t seconds after an object is shot directly upwards from the surface of some planet its height in feet is given by h=200t -2.5t^2.

a.) find a formula for the velocity of the bullet at time t.

b.) what is the maximum height attained by the bullet?

please show your work , it will make me easier to understand..:)
thank you so much and have fun

回答 (1)

Prof. Physics

2010-03-05 5:49 pm

✔ 最佳答案

1. x^2 + 4xy + y^2 = 13

Differentiate both sides with respect to x

2x + 4y + 4xdy/dx + 2ydy/dx = 0

(x + 2y) + (2x + y)dy/dx = 0

dy/dx = -(x + 2y)/(2x + y)

2. y = tan^-1(e^6x)

dy/dx = 6e^6x/[1 + (e^6x)^2] = 6e^6x / (1 + e^12x)

For y-intercept, put x = 0

dy/dx = 6/(1 + 1) = 3

Put x = 0, y = tan^-1(1) = pi/4

So, the equation of the line:

(y - pi/4)/(x - 0) = 3

3x = y - pi/4

3x - y + pi/4 = 0

3. f(x) = ln(x^5e^3x(x^4 + 3)^6)

f'(x) = 1/((x^5e^3x(x^4 + 3)^6) {[x^5e^3x 6(x^4 + 3)^5(4x^3)] + [(x^4 + 3)^6 (5x^4e^3x + 3x^5e^3x)}

= {x^4e^3x(x^4 + 3)^5[(24x^4) + (5 + 3x)(x^4 + 3)}/((x^5e^3x(x^4 + 3)^6)

= (24x^4 + (5 + 3x)(x^4 + 3))/x(x^4 + 3)

= (3x^5 + 29x^4 + 9x + 15) / (x^5 + 3x)

4.a. h = 200t - 2.5t^2

Velocity, v = dh/dt = 200 - 5t

b. d^2h/dt^2 = -5

Put dh/dt = 0, t = 40

WHen t = 40, d^2h/dt^2 = -5 < 0

So, maximum height is obtained when t = 40 s

Max. height = 4000m

參考: Physics king

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