Saturday

April 19, 2014

April 19, 2014

**Recent Homework Questions About Calculus**

Post a New Question | Current Questions

**Calculus**

sure you do. It's just a geometric series a = 1/e^2 r = 1/e^2 S = a/1-r = (1/e^2)/(1 - 1/e^2) = 1/(e^2-1)
*Thursday, April 3, 2014 at 7:50pm*

**Calculus**

20^n = 5^n * 4^n so, you have 5^n/20^n + 4^n/20^n = 1/4^n + 1/5^n both of which converge
*Thursday, April 3, 2014 at 7:44pm*

**Calculus**

Determine whether the series is convergent or divergent. If it is convergent, find its sum. If it is divergent, enter NONE. sum from 1 to infinity of (5^n+4^n)/20^n It is convergent, but I do not know how to find the sum.
*Thursday, April 3, 2014 at 1:15pm*

**Calculus**

Determine whether the series is convergent or divergent. If it is convergent, find its sum. If it is divergent, enter NONE. sum from 1 to infinity of 1/e^2n. It is convergent, but I do not know how to solve for the sum.
*Thursday, April 3, 2014 at 1:13pm*

**Calculus 12 Optimization**

43
*Thursday, April 3, 2014 at 7:19am*

**calculus**

I have this question in my homework .its a multiple choice, Select the correct answer. Answers are in billions of dollars per year. and the answers are a)137.4 b)44.4 c)22.2 d)16.96 e)27.48
*Thursday, April 3, 2014 at 4:02am*

**Calculus**

dy/dx
*Thursday, April 3, 2014 at 12:28am*

**calculus**

dg
*Wednesday, April 2, 2014 at 10:10pm*

**Calculus**

If y = y(x), write y’ in Leibniz notation.
*Wednesday, April 2, 2014 at 10:00pm*

**calculus**

let the two equal width be x let the single side by y xy = 600 y = 600/x cost = 20y + 30x = 20(600/x) + 30x d(cost)/dx = -12000/x^2 + 30 = 0 for a min of cost 30 = 12000/x^2 x^2 = 400 x = 20 the two equal sides are 20 yds each, and the long side is 600/20 = 30 yds.
*Wednesday, April 2, 2014 at 8:49pm*

**calculus**

revenue function of x^2+3p=675
*Wednesday, April 2, 2014 at 8:37pm*

**calculus**

a farmer wishes to fence off a rectangular plot of land, using an existing wall as one of the sides . the total are enclosed must be 600 square yards. the fence on the side parallel to the wall will cost 20$ per yard, while the fences on the other side will cost 30$ per yard. ...
*Wednesday, April 2, 2014 at 8:36pm*

**Calculus**

g'(x)=af'(x). Try to prove this using the definition of the derivative.
*Wednesday, April 2, 2014 at 8:29pm*

**Calculus**

If g(x) = a f(x); find g’(x). Is it g'(x)=af?
*Wednesday, April 2, 2014 at 8:13pm*

**calculus**

I see that C(1990) is given as 271.9 What are you asking?
*Wednesday, April 2, 2014 at 7:27pm*

**Calculus Help Please!!! Check**

looks like you are using Newton's Formula for Cooling properly.
*Wednesday, April 2, 2014 at 7:14pm*

**calculus**

Let C(t) be the total value of US currency (coins and banknotes) in circulation at time. The table gives values of this function from 1980 to 2000, as of September 30, in billions of dollars. Estimate the value of C(1990) . t 1980 1985 1990 1995 2000 C(t) 129.9 187.3 271.9 409...
*Wednesday, April 2, 2014 at 6:59pm*

**Calculus Help Please**

the linear app. appears to be best for the function ___f___ since it is closer to ___f__ for a larger domain than it is to _g and h____ . the app. looks worst for ___h__ since _h____ moves away from L faster than __f and g___ do.
*Wednesday, April 2, 2014 at 6:58pm*

**Calculus Help Please Check!!!**

Lf(x)= 1-2x Lg(x)= 1-2x Lh(x)= 1-2x the linear app. appears to be best for the function ___f___ since it is closer to _____ for a larger domain than it is to _____ . the app. looks worst for ___h__ since _____ moves away from L faster than _____ do. i couldn't figure out ...
*Wednesday, April 2, 2014 at 6:27pm*

**Calculus Help Please!!!**

I got the answer. It is 4 % increases in blood flow. thanks anyway.
*Wednesday, April 2, 2014 at 6:08pm*

**Calculus Help Please!!! Check **

When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round the answers to two decimal place.) (a) What is the temperature of the drink after 60 minutes? (b) When will its ...
*Wednesday, April 2, 2014 at 6:00pm*

**Calculus Help Please**

let f(x)=(x-1)^2, g(x)=e^(-2x), and h(x)=1+ ln(1-2x) find linearization of f, g, and h at a=0 Lf(x)= Lg(x)= Lh(x)= graph f,g, and h, and their linear app. For which function is the linear app. best? for which is the worst? explain. the linear app. appears to be best for the ...
*Wednesday, April 2, 2014 at 5:28pm*

**Calculus Help Please!!!**

When blood flows along a blood vessel, the flux F (the volume of blood per unit time that flows past a given point) is proportional to the fourth power of the radius R of the blood vessel: F = kR4 (This is known as Poiseuille's Law.) A partially clogged artery can be ...
*Wednesday, April 2, 2014 at 5:22pm*

**Calculus Help**

I can add more rats!! last step: dx/dt = -6( csc^2 (π/3)) (-π/3) = -6((4/3)(-π/3) = 24π/9 km/min = 8π/3 km/min (which is Steve's answer)
*Wednesday, April 2, 2014 at 5:10pm*

**Calculus Help**

At a time of t min after the plane passed over the tracking telescope, let the horizontal distance be x km Let the angle of elevation be Ø Make a sketch, I get cot Ø = x/6 x = 6cotØ dx/dt = -6 csc^2 Ø dØ/dt when Ø = π/3 and d&...
*Wednesday, April 2, 2014 at 5:06pm*

**Calculus - oops**

rats - that's dx/dt = -6csc^2θ dθ/dt when θ=π/3, we have x = 6/√3 and cscθ = 2/√ dx/dt = -6(4/3)(-π/3) = 8π/3 km/min
*Wednesday, April 2, 2014 at 5:02pm*

**Calculus Help**

At t=0, let x=0, so that at time t, x/6 = cotθ x = 6cotθ dx/dt = -6sec^2θ dθ/dt when θ=π/3, we have x = 6/√3 and secθ = 2 dx/dt = -6(2)(-π/3) = 4π km/min
*Wednesday, April 2, 2014 at 5:00pm*

**Calculus Help Please!!! Check**

Odd notation. P is a point (x,y) so usually we would see dy/dx = 3y y = c*e^3x 2 = c, so y = 2e^(3x) You divided the right side by 3 and multiplied the left aside by 3, but that was not correct: dy/dp = 3y ∫dy/y = ∫3dp ...
*Wednesday, April 2, 2014 at 4:50pm*

**Calculus Help**

A plane flies horizontally at an altitude of 6 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/3, this angle is decreasing at a rate of π/3 rad/min. How fast is the plane traveling at that time?
*Wednesday, April 2, 2014 at 4:47pm*

**calculus help thanks!**

PV = nRT V dP/dt + P dV/dt = nR dT/dt Now just plug in the data: (14)(0.13) + (8.0)(-0.17) = (10)(0.0821) dT/dt dT/dt = 0.56 °K/s
*Wednesday, April 2, 2014 at 4:45pm*

**Calculus Help Please!!!**

When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round the answers to two decimal place.) (a) What is the temperature of the drink after 60 minutes? (b) When will its ...
*Wednesday, April 2, 2014 at 4:43pm*

**Calculus Help Please!!! Check **

A curve passes through the point (0, 2) and has the property that the slope of the curve at every point P is three times the y-coordinate of P. Find an equation of the curve. dy/dp = 3y or ∫ (3/y) dy = ∫ dp or 3 ln(y) = p + c or @ (0,2) ln(2) = 0 + c or c = ln(2) 3...
*Wednesday, April 2, 2014 at 4:39pm*

**Calculus Help Please!!!**

given : dr/dt = 40 cm/s A = πr^2 dA/dt = 2π r dr/dt case 1: when s = 1, r = 40 dA/dt = 2π(40)(40) = 3200π cm^2/sec case 2: when s = 3, r = 3(4) = 120 cm dA/dt = 2π(120)(40) = 9600π cm^2/sec case 3: ... your turn .....
*Wednesday, April 2, 2014 at 4:37pm*

**calculus help thanks!**

The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at a certain instant, P = 8.0 atm and is ...
*Wednesday, April 2, 2014 at 4:35pm*

**Calculus Help Please!!!**

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 40 cm/s. Find the rate at which the area within the circle is increasing after each of the following. after 1 s = after 3 s = after 7 s =
*Wednesday, April 2, 2014 at 4:32pm*

**Calculus Help Please!!!**

f(x) is differentiable on the interval. So, you know there is a c such that f'(c) = (f(2)-f(0)/(2-0) = (-1-2)/2 = -3/2 So, just solve for c in 4c-5 = -3/2 c = 7/8 as a check, view http://www.wolframalpha.com/input/?i=plot+y+%3D+2x^2-5x%2B1+and+y+%3D+%28-3%2F2%...
*Tuesday, April 1, 2014 at 11:46pm*

**Calculus Help Please!!!**

well, f(x) is only defined on 0 <= x <= 1 Visit http://rechneronline.de/function-graphs/ and enter 8x*sqr(x-x^2) for your function. Set the range of y to be 2.5 to 3 and the peak will be quite clear
*Tuesday, April 1, 2014 at 11:40pm*

**Calculus Help Please!!!**

Consider the following. f(x) = 8x (square root of (x − x^2)) (a) Use a graph to find the absolute maximum and minimum values of the function to two decimal places.
*Tuesday, April 1, 2014 at 10:27pm*

**Calculus Help Please!!!**

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 2x^2 − 5x + 1, [0, 2] If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list...
*Tuesday, April 1, 2014 at 10:08pm*

**Calculus Help Please!!!**

done http://www.jiskha.com/display.cgi?id=1396400942 are you switching names???
*Tuesday, April 1, 2014 at 9:42pm*

**Calculus Help Please!!!**

Use the product rule to get f ' (t) = -(t^2 - 32)/√(64-t^2) = 0 for a local max/min t^2 = 32 t = ±4√2 f(±4√2) = 0 f(-1) = -1√63 f(8) = 8(√0) = 0 there you go, you got your f(t)'s what do you think? Here is your graph, notice ...
*Tuesday, April 1, 2014 at 9:41pm*

**Calculus Help!**

see http://www.jiskha.com/display.cgi?id=1396388805
*Tuesday, April 1, 2014 at 9:27pm*

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = xe^((−x^2)/98), [−6, 14]
*Tuesday, April 1, 2014 at 9:14pm*

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = (t square root of (64 − t^2)) ,[−1,8]
*Tuesday, April 1, 2014 at 9:11pm*

**Calculus Help!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = xe^(−x^(2)/98), [−6, 14]
*Tuesday, April 1, 2014 at 9:09pm*

**calculus**

5x^2
*Tuesday, April 1, 2014 at 8:19pm*

**Calculus Help Please!!!**

looks like we both messed up in the f(x) calculations should be f(4) = 2(64) - 3(16) - 72(4) + 7 = -201 f(-3) = 2(-27) - 3(9) - 72(-3)+7 = 142 end-values: f(-4) = 2(-64) - 3(16)( - 72(-4) + 7 = 119 f(5) = 2(125) - 3(25) - 72(5) + 7 = -178
*Tuesday, April 1, 2014 at 6:34pm*

**Calculus**

f ' (t) = -8sin t + 2(4 cos t) = -8sint + 8cost = 0 for a max/min 8sint = 8cost sint/cost = 1 tant = 1 t = 45° or 225° or t = π/4 , t = 5π/4 -->(outside our domain) so evaluate f(0) f(π/4) f(π/2) and determine which is the largest and which ...
*Tuesday, April 1, 2014 at 6:24pm*

**Calculus Help Please!!!**

so it the same way you were just given in your previous post. Let us know what you got
*Tuesday, April 1, 2014 at 6:07pm*

**Calculus Help Please!!!**

f ' (x) = 6x^2 - 6x - 72 = 0 for a local max/min x^2 - x - 12 = 0 (x-4)(x+3) = 0 x = 4, or x = -3 f(4) = 2(64) - 3(16) - 72(4) + 7 = -201 f(-3) = 2(-27) - 3(9) - 72(-3)+7 = 115 end-values: f(-4) = 2(-64) - 2(16)( - 72(-4) + 7 = 135 f(5) = 2(125) - 3(25) - 72(5) + 7 = -178 ...
*Tuesday, April 1, 2014 at 6:04pm*

**Calculus Help Please!!!**

f'(x) = 6x^2 -6x -72 f'(0) = 6(x^2 -x-12) 0 = 6(x+ 3) (x-4) x = -3, 4 f(-4) = 2(-4)^3 -3(-4)^2 -72(-4) +7= 139 f(-3) = 2(-3)^3 -3(-3)^2 -72(-3) +7=132 f(4) = 2(4)^3 -3(4)^2 -72(4) + 7= -145 f(5) = 2(5)^3 -3(5)^2 -72(5)+7= -178
*Tuesday, April 1, 2014 at 6:01pm*

**Calculus**

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 8 cos t + 4 sin 2t, [0, π/2]
*Tuesday, April 1, 2014 at 5:45pm*

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = (t square root of (64 − t^2)), [−1, 8]
*Tuesday, April 1, 2014 at 5:27pm*

**Calculus Help Please!!!**

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 2x^3 − 3x^2 − 72x + 7 , [−4, 5]
*Tuesday, April 1, 2014 at 5:26pm*

**calculus and vectors**

Add the 2 equations ... 6x + y = 6 y = 6-6x let x = 0 , y = 6 2(0) + 4(6) + z = 5 --> z = -19) we have a point (0,6,-19) on the line of intersection let x = 1, y = 0 2(1) + 4(0) + z = 5 --> z = 3 and (1,0,3) is another point on that line So a direction vector of that ...
*Tuesday, April 1, 2014 at 5:18pm*

**calculus and vectors**

Determine an equation of the line of intersection of the planes 4x − 3y − z = 1 and 2x + 4y + z = 5.
*Tuesday, April 1, 2014 at 4:59pm*

**calculus**

max area for a given perimeter is a circle. Is the wire evenly divided? If so, c = p/2, so r = p/4π, and the total area is a = 2(πr^2) = 2π(p^2/16π^2) = p^2/8π If not, then if we have a tiny circle, (area effectively zero), then r = p/2π and a...
*Tuesday, April 1, 2014 at 12:30pm*

**calculus**

Twenty feet wire is used to make two figures? What is the maximum areas of enclosed figures.
*Tuesday, April 1, 2014 at 11:14am*

**Calculus**

Thanks Mr. Steve for the guidance.
*Tuesday, April 1, 2014 at 5:21am*

**Calculus**

Looks good to me. I get ∫[1,2] (3-x) - 2/x dx = 3x - 1/2 x^2 - 2logx [1,2] = 3/2 - 2log2 gotta be a typo in the answer. 2log2 is correct.
*Tuesday, April 1, 2014 at 4:52am*

**Calculus**

Find the area cut off by x+y=3 from xy=2. I have proceeded as under: y=x/2. Substituting this value we get x+x/2=3 Or x+x/2-3=0 Or x^2-3x+2=0 Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the line x+y=3. Area under curve above X axis ...
*Tuesday, April 1, 2014 at 3:07am*

**Calculus**

Good idea. However, in an attempt to use your math, and also to apply to a possibly more general problem later, try implicit differentiation: x^2 + y^2 = 18 2x + 2yy' = 0 y' = -x/y So, if -x/y = 1 and x^2+y^2 = 18, 2x^2 = 18 x = ±3 Now, figuring y should not be ...
*Monday, March 31, 2014 at 1:02pm*

**Calculus**

LOL, sketch a graph !
*Monday, March 31, 2014 at 11:46am*

**Calculus**

The circle defined by the equation x^2 + y^2 = 18 has two points where the slope of its tangent line is m=1. Find those points.
*Monday, March 31, 2014 at 11:41am*

**Calculus: Integral**

recall that sec^2 = 1+tan^2, so you have ∫sec^4(4x) dx = ∫sec^2(4x)(1 + tan^2(4x)) dx = ∫sec^2(4x) dx + ∫tan^2(4x) sec^2(4x) dx = 1/4 tan(4x) + (1/4)(1/3) tan^3(4x) and you can massage that in several ways.
*Monday, March 31, 2014 at 5:32am*

**Calculus: Integral**

I don't understand how to do this one integral problem that involves secant. I'm asked to find the integral of sec^4 (4x). I'm not really sure how to go about solving this problem.
*Monday, March 31, 2014 at 3:32am*

**Calculus Help Please!!!**

looking at a diagram, if A is a away from Q and B is b away from Q, then √(a^2+144) + √(b^2+144) = 39 a/√(a^2+144) da/dt + b/√(b^2+144) db/dt = 0 Now just plug in da/dt = 3.5 a = 5 b = 23.065 (from 1st equation when a=5) and solve for db/dt
*Monday, March 31, 2014 at 12:27am*

**Calculus Help**

v = 2/3 pi r^3 (half a sphere) dv = 2 pi r^2 dr now just plug in the given r and dr watch the units.
*Monday, March 31, 2014 at 12:07am*

**Calculus Help Please!!!**

v = 4/3 pi r^3 dv = 4 pi r^2 dr c = 2pi r dc = 2pi dr so, dr = dc/(2pi) meaning that dv = 4 pi r^2 dc/(2pi) = 2 r^2 dr so, using the given numbers, dv = 2*(80/2pi)^2 * 0.5 = 1600/pi^2
*Monday, March 31, 2014 at 12:05am*

**Calculus Help Please!!!**

looks good to me
*Monday, March 31, 2014 at 12:01am*

**pre calculus**

24
*Sunday, March 30, 2014 at 11:40pm*

**Calculus Help Please!!!**

Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.) fourth root of (1 + 2x)≈ 1 + (1/2)x
*Sunday, March 30, 2014 at 11:21pm*

**Calculus Help Please!!!**

The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. 1) Use differentials to estimate the maximum error in the calculated volume. (Round your answer to the nearest integer.) 2) What is the relative error? (Round your answer to three decimal ...
*Sunday, March 30, 2014 at 11:17pm*

**Calculus Help**

Use differentials to estimate the amount of paint needed to apply a coat of paint 0.03 cm thick to a hemispherical dome with diameter 54 m. (Round your answer to two decimal places.)
*Sunday, March 30, 2014 at 11:16pm*

**Calculus Help Please**

f(x) = f(a) + (x-a) f'(a) f(x) = x^4 df/dx = 4x^3 let a = 2 then f(a) = 2^4 = 16 f'(a) = 4*8 = 32 f(x) = 16 + (x-a)(32) x-a = - .001 so f(1.999) = 16 -.001(32) = 16 - .032 f(1.999) = 15.968 with calculator it is 15.968 also
*Sunday, March 30, 2014 at 10:29pm*

**Calculus Help Please!!!**

at x = 3 y = 3*3 - 9 = 0 at x = 2.4 y = 3(2.4) - 2.4^2 = 1.44 delta y = 1.44 -0 = 1.44 dy/dx = 3 - 2 x at x = 3 dy/dx = 3 - 6 = -3 if dx = -.6 dy = -3 (-.6) = 1.8
*Sunday, March 30, 2014 at 10:21pm*

**Calculus**

Thanks Damon, that really clears it up for me
*Sunday, March 30, 2014 at 10:17pm*

**Calculus Help Please**

Use a linear approximation (or differentials) to estimate the given number. (1.999)^4
*Sunday, March 30, 2014 at 10:14pm*

**Calculus Help Please!!!**

Oh, I see f(x) = ln (1+x) df/dx = 1/(1+x) d^2f/dx^2 = -1/(1+x)^2 f(x) = f(0) + [1/(1+0)] x - x^2/2! ... f(x) = 0 + x - x^2/2 + ..... well at a first cut when is x^2/2 =.1 x? x/2 = .1 x = .2
*Sunday, March 30, 2014 at 10:13pm*

**Calculus Help Please!!!**

Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) y = 3x − x^2, x = 3, Δx = −0.6 Δy=???
*Sunday, March 30, 2014 at 9:57pm*

**Calculus Help Please!!!**

Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.) ln(1 + x) ≈ x xE
*Sunday, March 30, 2014 at 8:28pm*

**Calculus Help Please!!!**

Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the figure). The point Q is on the floor h = 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 3.5 ft/s. How fast is cart B moving toward Q at...
*Sunday, March 30, 2014 at 7:16pm*

**Calculus Help Please!!!**

LOL - Guess which of us is the mathematician and which is the Engineer :)
*Sunday, March 30, 2014 at 7:15pm*

**Calculus Help Please!!!**

when the water has depth x, the cross-section is a trapezoid with bases 30 and 30+x. So the volume of water at depth x is v = (60+x)/2 * x * 500 cm^3 = 250x^2 + 15000x so, knowing that dv/dt = (500x + 15000) dx/dt just solve that for dx/dt when x=20
*Sunday, March 30, 2014 at 7:12pm*

**Calculus Help Please!!!**

Q = incoming flow rate = .1 m^3/min dh/ dt = Q A where A = surface area = length * width at 20 cm depth which depth is (1/2) height width = 30 + 1/2(70-30) = 30+20 = 50 cm = .5 m wide water surface so A = 5 * .50 =2.5 m^3 so finally dh/dt = .1 * 2.5 = .25 m/min = 25 cm/min
*Sunday, March 30, 2014 at 7:11pm*

**Calculus Help Please!!!**

A water trough is 5 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 70 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.1 m3/min how fast is the water level rising when ...
*Sunday, March 30, 2014 at 6:43pm*

**physics**

find the difference in PE in the two locations. Hold one charge stationary. PEtotal= kQQ/.1 PE(new total)=KQQ/.06 subtract the first from the second, that must equal the work done. There are more difficult ways to work this, involving finding work in an integral calculus ...
*Sunday, March 30, 2014 at 10:11am*

**Calculus**

when dy/du = u^½ what does y = ? Just a simple power rule substitution. dy/du = u^n y = u^(n+1) / (n+1) + C
*Saturday, March 29, 2014 at 4:38pm*

**Calculus**

Thanks Steve! Finally got this one correct on Math Lab! I am eternally grateful, and have made an account here so I can get help and I have already tried helping others as well. I am very pleased with the service of this site and am glad to have found it :) Hooray!!!
*Saturday, March 29, 2014 at 4:34pm*

**Calculus**

When dy/dx=(x-6)^½ what does y equal?
*Saturday, March 29, 2014 at 4:11pm*

**Calculus**

the rate of change of volume is the surface area times the rate of change of height 450 ft^3/min = surface area * dh/dt surface area = pi r^2 450 = pi (30^2) dh/dt dh/dt = .159 ft/min You could do this by saying V = pi r^2 h dV/dh = pi r^2 dV/dh*dh/dt = pi r^2 dh/dt chain rule...
*Friday, March 28, 2014 at 11:48pm*

**Calculus**

Jesse has constructed a huge cylindrical can with a diameter of 60 ft. The can is being filled with water at a rate of 450 ft3/min. How fast is the depth of the water increasing? (Hint: The volume of water in the cylinder is determined by πr2h where r is the radius and h ...
*Friday, March 28, 2014 at 11:39pm*

**math**

assume no calculus allowed thus complete he square to find the vertex of this parabola 16 t^2 - 20 t - 2 = -h 16 t^2 -20 t = - h + 2 t^2 - 5/4 t = - h/16 + 1/8 t^2- 5/4 t+ 25/64 = -h/16 + 8/64 + 25/64 (t - 5/8)^2 = -(1/16)(h - 33/4) so in 5/8 of a second it reaches the vertex ...
*Friday, March 28, 2014 at 10:37pm*

**Calculus**

so it is positive for x>.316 and for x <.316
*Friday, March 28, 2014 at 8:12pm*

**Calculus**

Note that you did the second derivative correctly. It is easier to write it their way y" = 10e^(-5x^2)(10x^2-1) It is zero at x = +/- .316
*Friday, March 28, 2014 at 8:11pm*

**Calculus**

first http://www.wolframalpha.com/input/?i=plot++e^%28-5x^2%29 Then this for second derivative and graph https://www.wolframalpha.com/input/?i=second+derivative+of+e^%28-5x^2%29
*Friday, March 28, 2014 at 7:57pm*

**Calculus**

first http://www.wolframalpha.com/input/?i=plot++e^%28-5x^2%29
*Friday, March 28, 2014 at 7:52pm*

**Calculus**

At what interval is e^(-5x^2) concave up? I know the second derivative is 100x^2*e^-5x^2-10*e^-5x^2 but I just can not figure this one out. Thank you for your help!
*Friday, March 28, 2014 at 7:38pm*

**Calculus**

Thanks for the advice. I checked the problem statment and answer several times and got the same result. I also suspect it to be a print mistake in the book.
*Friday, March 28, 2014 at 7:46am*

**Calculus**

As a first check, I went to http://www.wolframalpha.com/input/?i=2%E2%88%AB[3%2C4]+%282%2F3+%E2%88%9A%28x^2-9%29%29+dx and saw that they show the area as 2.28 So, I suspect there is an error in the problem or the answer. Your calculation appears to be correct, ...
*Friday, March 28, 2014 at 5:43am*

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