1. The altitude of a triangle is increasing at a rate of 3 cm/min while the area
ID: 2893405 • Letter: 1
Question
1. The altitude of a triangle is increasing at a rate of 3 cm/min while the area of the triangle is increasing at a rate of 2.5 square cm/min. At what rate is the base of the triangle changing when the altitude of 11.5 cm and the area is 89 sq cm?
2. Gravel is being dumped from a conveyor belt at a rate of 10 cubic ft. per min. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 6 ft high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V=1/3pi(r)^2h
? ft/min
3. A police car is located 70 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 90 feet up the road from the location of the police car.The police radar reads that the distance between the police car and the red car is decreasing at a rate of 80 feet. per second. How fast is the red car actually traveling along the road? The actual speed (along the road) of the red car is ? feet per second
please show work and dont round your final answer thank you
Explanation / Answer
1)
for a triangle,let altitude =h , base =b , area =A
area,A=(1/2)bh
given altitude is 11.5 cm and the area is 89 sq cm
h=11.5, A=89
=>89=(1/2)*b*11.5
=>b=178/11.5
=>b=1780/115
A=(1/2)bh
differentiate with respect to time t
=>dA/dt=(1/2)[((db/dt)*(h))+((b)*(dh/dt))]
altitude of a triangle is increasing at a rate of 3 cm/min , area of the triangle is increasing at a rate of 2.5 square cm/min =>dh/dt =3 ,dA/dt =2.5
=>2.5=(1/2)[((db/dt)*11.5)+((1780/115)*(3))]
=>[((db/dt)*11.5)+((1780/115)*(3))] =5
=>((db/dt)*11.5) =5-((1780/115)*(3))
=>(db/dt) =(1/11.5)*[5-((1780/115)*(3))]
=>(db/dt) =-1906/529-3.603
base of the triangle is changing(decreasing) at 1906/5293.603 cm/min
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