Mth 231 written assignment 4. PLEASE TYPE IN MATH LAB OR MATH TYPE AND RE STATE THE PROBLEM BEFORE WRITING ANSWER. PLEASE RE ANSWER THE QUESTION WITH RIGHT ANSWER.
The problem and instructions need to be stated, more points taken off if problem and instructions are not stated next time
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Houstin Claiborne Houstin Claiborne 0586348 MTH231 Module 8—Written Assignment 4 1. Z^2 = x^2 + y^2, at (1, 3), z = √(1)^2 + (3)^2 =√10 2(√10)dz/dt = 2(1)dx/dt+2(3)dy/dt=2(4)+6(3)=26 Dz/dt = 26/2√10=13√10/10 (dz)/(dt)=(13√10)/(10) 0. You and a friend are riding your bikes to a restaurant that you think is east; your friend thinks the restaurant is north. You both leave from the same point, with you riding at 16 mph east and your friend riding 12 mph north. After you traveled 4 mi, at what rate is the distance between you changing? 5(H’(t))=3(12)+4(16) 5H’(t) = 36+64 5H’(t)=100 H’(t)=100/5 H’(t)=20 The distance changes at the rate of 20mph after 4 miles 0. Consider a right cone that is leaking water. The dimensions of the conical tank are a height of 16 ft and a radius of 5 ft. How fast does the depth of the water change when the water is 10 ft high if the cone leaks water at a rate of 10 ft3/min? V’(t)=25/768π3h’(t)(h(t))^2=25/256πh’(t)(h(t))^2 h’(t)=256/25π(h(t))^2V’(t) h’(t)=256/25π(10)^2x10 h’(t)=256/250π=128/125πft/min 0. A tank is shaped like an upside-down square pyramid, with base of 4 m by 4 m and a height of 12 m (see the following figure). How fast does the height increase when the water is 2 m deep if water is being pumped in at a rate of 2/3 m3/sec? a=1/3h Volume=1/27h^3 dV/dt=(h^2/9)(dh/dt) dh/dt=6h^2 = 6(2)^2 = 24m/sec The rate of height of pyramid at which rising in it is 24m/min when the height is 2m 0. Determine where the local and absolute maxima and minima occur on the graph given.Assume the graph represents the entirety of each function. The absolute minimum occurs at x=-2, x=2 and the absolute maximum at x=-2.5, x=2.5 and the local minima occurs at x=0, local maxima is at x=-1, x=-1 0. Find the local and absolute minima and maxima for the function over (−∞, ∞). The local minima of function y(x) over the interval (-Ꝏ, Ꝏ) is 0 at x=0 and x=1, the local maxima of the function y(x) over the interval (-Ꝏ, Ꝏ) is 0.002 at x=0.5 0. Use the Mean Value Theorem and find all points 0 < c="">< 2 such that f (2) − f (0) = f ′(c)(2 − 0). f’(c)(2-0)=f(2)-f(0) (c-1)^9=0 c-1=0 c=1 0. analyze the graphs of f ′, then list all intervals where f is increasing or decreasing. the function is increasing in (-1, 0) u (0, 1) the function is decreasing in (-2, -1) u (1, 2) 0. analyze the graphs of f ′, then list all inflection points and intervals f that are concave up and concave down. there is one point of inflection at x=1 and it is concave up between -1 and 0 0. for the following function: f(x) = x3 − 6x2, determine a. intervals where f is increasing or decreasing, f(x) is increasing in (-ꝏ, 0) and (4, ꝏ) f(x) is decreasing in interval (0, 4) b. local minima and maxima of f, local minima is at x=4 and local maxima is at x=0 c. intervals where f is concave up and concave down, and f(x) is concave up in the interval (2, ꝏ) and concave down in the interval (-ꝏ, 2) d. the inflection points of f. (2, -16) is the point of inflection 0. evaluate the following limit: the limit of this function is 4. 0. evaluate the following limit: the limit of this function is 4. 0. find the horizontal and vertical asymptotes for the vertical asymptotes are x=1 and x=-1 the horizontal asymptote is y=0 0. find the horizontal and vertical asymptotes for the vertical asymptotes are x=2 and x=-2 there is no horizontal asymptote of this function 0. set up and evaluate the following optimization problem: to carry a suitcase on an airplane, the length + width + height of the box must be less than or equal to 62 in. assuming the height is fixed, show that the maximum volume is . what height allows you to have the largest volume? the maxima volume for a fixed height h will be given by v = h(31-(1/2)h)^2 at height 20.667 in, volume is the largest. 0. draw the given optimization problem and solve: find the dimensions of the closed cylinder volume v = 16π that has the least amount of surface area. the dimensions of the closed cylinder volume v=16π that has the least amount of surface area are radius is 2 units and the height is 4 units 0. limx→a((x-a)/(x^2-a^2))=1/2a where a is not equal to 0 0. evaluate the limit limx→0((sinx-tanx)/x^3)=-1/2 0. evaluate the limit limx→ꝏ((sinx-x)/x^2) = 0 0. evaluate the limit limx→π/4(1-tanx)cotx=-1 2="" such="" that ="" f="" (2)="" −="" f="" (0)="f" ′(c)(2="" −="" 0).="" f’(c)(2-0)="f(2)-f(0)" (c-1)^9="0" c-1="0" c="1" 0.="" analyze="" the="" graphs="" of="" f="" ′,="" then="" list="" all="" intervals="" where="" f="" is="" increasing="" or="" decreasing.="" the="" function="" is="" increasing="" in="" (-1,="" 0)="" u="" (0,="" 1)="" the="" function="" is="" decreasing="" in="" (-2,="" -1)="" u="" (1,="" 2)="" 0.="" analyze="" the="" graphs="" of="" f="" ′,="" then="" list="" all="" inflection="" points="" and="" intervals="" f="" that="" are="" concave="" up="" and="" concave="" down.="" there="" is="" one="" point="" of="" inflection="" at="" x="1" and="" it="" is="" concave="" up="" between="" -1="" and="" 0="" 0.="" for="" the="" following="" function:="" f(x)="x3" −="" 6x2,="" determine="" a.="" intervals="" where="" f="" is="" increasing="" or="" decreasing,="" f(x)="" is="" increasing="" in="" (-ꝏ,="" 0)="" and="" (4,="" ꝏ)="" f(x)="" is="" decreasing="" in="" interval="" (0,="" 4)="" b.="" local="" minima="" and="" maxima="" of="" f,="" local="" minima="" is="" at="" x="4" and="" local="" maxima="" is="" at="" x="0" c.="" intervals="" where="" f="" is="" concave="" up="" and="" concave="" down,="" and="" f(x)="" is="" concave="" up="" in="" the="" interval="" (2,="" ꝏ)="" and="" concave="" down="" in="" the="" interval="" (-ꝏ,="" 2)="" d.="" the="" inflection="" points="" of="" f.="" (2,="" -16)="" is="" the="" point="" of="" inflection="" 0.="" evaluate="" the="" following="" limit:="" the="" limit="" of="" this="" function="" is="" 4.="" 0.="" evaluate="" the="" following="" limit:="" the="" limit="" of="" this="" function="" is="" 4.="" 0.="" find="" the="" horizontal="" and="" vertical="" asymptotes="" for="" the="" vertical="" asymptotes="" are="" x="1" and="" x="-1" the="" horizontal="" asymptote="" is="" y="0" 0.="" find="" the="" horizontal="" and="" vertical="" asymptotes="" for="" the="" vertical="" asymptotes="" are="" x="2" and="" x="-2" there="" is="" no="" horizontal="" asymptote="" of="" this="" function="" 0.="" set="" up="" and="" evaluate="" the="" following="" optimization="" problem:="" to="" carry="" a="" suitcase="" on="" an="" airplane,="" the="" length="" +="" width="" +="" height="" of="" the="" box="" must="" be="" less="" than="" or="" equal="" to="" 62="" in.="" assuming="" the="" height="" is="" fixed,="" show="" that="" the="" maximum="" volume="" is="" .="" what="" height="" allows="" you="" to="" have="" the="" largest="" volume?="" the="" maxima="" volume="" for="" a="" fixed="" height="" h="" will="" be="" given="" by="" v="h(31-(1/2)h)^2" at="" height="" 20.667="" in,="" volume="" is="" the="" largest.="" 0.="" draw="" the="" given="" optimization="" problem="" and="" solve:="" find="" the="" dimensions="" of="" the="" closed="" cylinder="" volume="" v="16π" that="" has="" the="" least="" amount="" of="" surface="" area.="" the="" dimensions="" of="" the="" closed="" cylinder="" volume="" v="16π" that="" has="" the="" least="" amount="" of="" surface="" area="" are="" radius="" is="" 2="" units="" and="" the="" height="" is="" 4="" units="" 0.="" limx→a((x-a)/(x^2-a^2))="1/2a" where="" a="" is="" not="" equal="" to="" 0="" 0.="" evaluate="" the="" limit="" limx→0((sinx-tanx)/x^3)="-1/2" 0.="" evaluate="" the="" limit="" limx→ꝏ((sinx-x)/x^2)="0" 0.="" evaluate="" the="" limit="" limx→π/4(1-tanx)cotx="">