Please help me with my last calculus question?
A police cruiser hunting for a suspect pulls over and stops at a point 20 ft from a straight wall. The flasher on top of the cruiser revolves at a constant rate of 60 deg/sec, and the light beam casts a spot of light as it strikes the wall. How fast is the spot of light moving along the wall at a point 50 ft from the point on the wall closest to the cruiser? (Round your answer to one decimal place.)
Need help with last calculus question!!!!!?
So we need a line through (-4,3) and (x0,0), that is tangent to the curve x - x^2, and the question is, what is the smallest value of x0 that is possible. The tangent line to x - x^2 will have a slope 1 - 2x. It must go through (-4,3) and (x,x - x^2), so its slope is (x - x^2 - 3)/(x+4). Hence we have: (1 - 2x)(x+4) = x - x^2 - 3 => 4 - 2x^2 - 7x = x - x^2 - 3 => 0 = x^2 + 8x - 7 => x = -4 + (1/2)*sqrt(64 + 28) = -4 + sqrt(16+7) = 0.7958. Hence the line is y = -0.5916x + (3 - 4*0.5916) = -0.5916x + 0.6336. Finally, y will be 0 where x = 0.6336/0.5916 = 1.071. The top of the hill is at (1/2, 1/4), and the closest the receiver can be is at (1.071,0), that is, 0.571 units to the right of the top of the hill. Please check my arithmetic!
Can anyone please help me on my last hw question ? Calculus?
Assuming the relationship is linear, the slope of the c.p.m. vs. temperature graph, (where c.p.m. is chirps per minute), is (173 - 112) / (90 - 65) = 61 / 25 = 2.44 c.p.m./(degree F). So with N as the number of c.p.m. and T as the temperature in degrees Fahrenheit, the linear model will have the form N - 112 = (2.44)*(T - 65) ----> N = (2.44)*T - (2.44)*65 + 112 ----> N = (2.44)*T - 46.6. Note that this linear model would be less reliable for extreme temperatures. For example, with T < 19 degrees F this model would give a negative value for N, and crickets can't 'un-chirp', as far as I know. :)
Very last calculus question!?
Ok, this is the last thing I need help with for the night. These are things that we learned in Calculus 1, but I don't remember how to do them... and unfortunately, I don't have my notes anymore. There are a few problems here with the answers, but I need an explination of how to actually arrive at the right answer. So, if you could even help me with only one that would be great! 1. Find area bounded by given curves: y = xe^-0.4, y = 0, and x = 5 Answer: (25/4) – (75/4)e^-2 2. Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the regi9on bounded by the curves. y = x sinx, y = (x-2)^2 Answer: 1.0475, 2.8731; 2.1828
Could you help with this calculus question? (MVT)
Thanks for A2A, Paul.The statement is not valid in this form.As David Joyce showed, one possibility to repair it is to assume [math]ab>0.[/math] Counter-example: [math] a=-1, b=1, f(x)=x^2-1[/math]. Since [math]f(1)=f(-1)=0[/math], it satisfies the condition above.If there had been such [math] c[/math], it would have held [math] f(1) +f(-1) - (c^2-1) =(1 + (-1) -c) 2c[/math]So [math]c^2-1=2c^2[/math] and thus [math]c^2=-1[/math]. A contradiction.Moreover, I don't know what is a "Rolle function", I know only Rolle's theorem.
Can you do a calculus question?
Yes I can (it may not be right though) :-)
Can someone please help with a calc question for my final?
To determine the amount of water flowing down a certain 100-yard-wide river, engineers need to know the area of a vertical cross section of the river. Measurements of the depth of the river were made every 20 yards from one bank to the other. The readings in fathoms were 0, 22, 33, 44, 22, 0. (One fathom equals 2 yards.) Use the trapezoidal rule to estimate the area of the cross section.
Minimum cost Calc Question? PLEASE HELP!?
Im on my last question and im stuck!. I need this question and don't even know how to start it: An industrial tank of the shape described in Exercise 39 must have a volume of 3000 cubic feet. The hemispherical ends cost twice as much per square foot of surface area as the sides. Find the dimensions that will minimize cost. It mentions question 39 so here: A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 12 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area. The radius is- r the equal sign but wavy 1.42
CALCULUS HOMEWORK QUESTION! PLEASE HELP!?
In m/s^2, the value of g is 9.8 9.8 m/s^2 * 100 cm/m * 1 in/2.54 cm * 1 ft/12 in = 980÷ 30.48 This is approximately 32 ft/s^2. 1. Determine the position and velocity functions for the coin. hf = hi – vi * t – ½ * g* t^2 Since the coin is dropped, its initial velocity is 0 ft/s. hf = 1454 – 16 * t^2 vf = 32 * t 2. Determine the average velocity of the coin on the interval [1, 3]. vf1 = 32 ft/s vf3 = 32 * 3 = 96 ft/s These are the answers to #3. Average = ½ * (32+ 96) = 64 ft/s. 4. At what time is the instantaneous velocity of the coin equal to the average velocity of the coin found in Part B? This is at the beginning. At that time, it velocity is 0 ft/s 5. What is the name of the theorem that says there must be at least one solution to Part D? I do not know. 6. Find the velocity of the coin just before it hits the ground. Let’s use the following equation to determine the time when the coin hits the ground hf = 1454 – 16 * t^2 hf = 0 0 = 1454 – 16 * t^2 t = √(1454 ÷ 16) = √90.875 This is approximately 9.23 seconds. vf = 32 * √(1454 ÷ 16) = √90.875 This is approximately 305 ft/s. Or In physics, we use the following equation. vf^2 = vi^2 + 2 * a * d, vi = 0 vf^2 = 2 * 32 * 1454 vf = √√90.875 This is approximately 305 ft/s. This is a good way to check the answer.
Last AP Calculus help?
let: X=length of one side of the square C=circumference of the circle P=perimeter of the square Y=area enclosed between the circle and the square a.) dP/dt=? dC/dt=6 in per sec solution: C=pi*X dC/dt=pi*dX/dt dx/dt=6/pi inches per sec P=4X dP/dt=4dX/dt=4*(6/pi) inches per sec thus, dP/dt=24/pi =7.64inches per sec b.) dY/dt=? when Area of circle is =25pi Y=Area of square - Area of circle Y=X^2-(pi*X^2)/4 Y=X^2(1-pi/4) dY/dt=(1-pi/4)*2X*dX/dt (eq 1) Area of circle = 25pi pi*r^2=25pi r=5 inches X=2r=10 (subs to eq 1) dY/dt=(1-pi/4)*2*10*6/pi thus, dY/dt=8.2 square inches per sec