TRENDING NEWS

POPULAR NEWS

I Need Help With This Calc Question Help

I need help on a calculus question...plz help..?

The decay equation for a radioactive substance is known to be y = y0e^-0.054t, with t in days. About how long will it take for the amount of substance to decay to 79% of its original value?

Im extremely confused... every time i try to take the anti derivative of the equation i forget what t do with the variable y0.. plz hlp if u can. is there another rule idk about? if u could show some steps and explain a lil... that would be awesome..

thank you!

Could you help with this calculus question?

[math]\int_{0}^{x_C} g(x)-f(x) dx[/math] where [math]g(x)[/math] is the funny curve ABC’s equation, [math]f(x)[/math] is the line with slope [math]\sqrt{3}[/math] (OC) and [math]x_C[/math] is the x coordinate of point C.That’s the easiest integral to set up and evaluate, you can think of summing vertical strips with width dx.

5 calculus questions help needed!?

I'm having a real hard time with these 5 questions ! help appreciated ! xoxo

1. Given a central angle of 76.4 degrees find the length of the itercepted arc in a circle of radius 6 centimeters. Round to nearest tenth.

2. Find the area of a sector if the central angle measures 7pi/12 radians and the radius of the circle is 2.6 meters. Round to the nearest tenth.

3. A belt runs a pully that has a diameter of 12 centimeters. If the pulley rotates at 80 revolutions per minute, what is its angular velocity in radians per second and its linear velocity in centimeters per second?

4. State the amplitude and the period for y=8/3 cos 6theta/5

5. Write an equation of the sine function with amplitude 5 and period 5pi/6

Calculus Limit Questions HELP NEEDED!!?

HELP IS GREATLY APPRECIATED THANK YOU!!

The lim as h->0 tan3(x+h) - tan3x / h
a) 0
b) 3sec^2(3x)
c) sec^2(3x)
3) 3cot(3x)
e) nonexistant


If f(x) = e^x, which of the following equals f '(e)??
a) lim as h->0 e^x+h / h
b) lim as h->0 e^x+h - e^e / h
c) lim as h->0 e^e+h - e / h
d) lim as h->0 e^x+h - 1 / h
e) lim as h->0 e^e+h - e^e / h


lim as h->0 ln(e+h) - 1 / h
a) f '(e), where f(x) = ln x
b) f '(e), where f(x) = ln x / x
c) f '(1), where f(x) = ln x
d) f '(1), where f(x) = ln (x+e)
e) f '(0), where f(x) = ln x
-----------

In addition to the answers, please explain how you solve em, thanks!!

I need help with this calculus question (accumulated change)?

You need to derive the equations for the distance traveled for both cars.

x = distance
S = integral symbol

Car A is easy: xa = v * t or xa = 40t
Car B's distance equation derivation requires some calculus.
dx=v(t)dt -> integrate both sides
xb (0(@2hrs the distance travelled = 90mi)
xb (t>2hrs) = 90 + Sv(t)dt (from 2 to t) = 90 + S 60-5tdt (from 2 to t) = 60t - 2.5t^2 (from 2 to t) = 90 + (60t - 2.5t^2) - (60 * 2 - 2.5 * 2^2) = 60t - 2.5t^2 - 20

Recap
xa = 40t
xb (0xb (t>2hrs) = -2.5t^2 + 60t - 20

A.
The distance that Car B is ahead is xb - xa. Car B will still be going faster than Car A after 2hrs so it will be getting further ahead for a while even though it will be slowing down at that point. We need to use the t>2hrs equation.

xb - xa = -2.5t^2 + 60t - 20 - 40t = -2.5t^2 + 20t - 20

We are looking for a maximum so we need to take the derivitive of this equation and set it equal to zero to know when the peak of the parabola has been reached.

-5t + 20 = 0
t = 4hrs

This could have been solved using common sense too but that often leads to errors. In this case common sense actually wins out. The point at which the maximum distance between the cars will be achieved is when Car B's velocity falls below that of Car A. They gave us the velocity equations. Car B's velocty goes up to 50mph after 2hrs and then at the same rate declines back to 40mph 2hrs later.

Anyway, it is always safer to calculate the answer.

B. xb(5)-xa(5) = -2.5 * 5^2 + 20 * 5 - 20 = 17.5mi

C. Essentially, at what time is xb - xa equal to 0.
-2.5t^2 + 20t - 20 = 0
t = 6.83hrs (1.17hrs can be ignored since it is in the 0 - 2hr time slot for which we would have to switch back to the first equation and work backwards)

t = 6.83hrs

The above process is correct but I didn't double check the individual numbers so there might be some small errors. Make sure to work through it.

I need help with a calculus question involving the product rule!?

I have the answer but can't seem to get it.
Please could you show me step by step on how you found the answer, that would be great!
answer:
-4.84L/h
Question:

A 75 L gas tank has a leak. After t hours, the remaining volume, V, in litres is V(t)= 75(1-t/24)^2.
0< t <24 . Use the product rule to determine how quickly the gas is leaking from the tank when the tank is 60% full of gas.

{[both ( < ) signs mean less than or equal too t and 24,]}

Can anyone help me find the other variable in this calculus question?

For the function to be continuous at x=2, we need to have,Left hand limit=right hand limit=functional value of function as variable tends to a limiting value(at the neighborhood of the limiting value). so f(2-)=f(2)=f(2+),so we have as variable x tending to 2,4a+6=2b+4=-2which implies that,a=-2 and 2b+4=-2which implies that(solving for b), we get b=-3.

TRENDING NEWS