Answer:
32.66 units
Step-by-step explanation:
We are given that
Point A=(-2,-4) and point B=(1,20)
Differentiate w.r. t x
We know that length of curve
We have a=-2 and b=1
Using the formula
Length of curve=
Using substitution method
Substitute t=12x+14
Differentiate w.r t. x
Length of curve=
We know that
By using the formula
Length of curve=![s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B12%7D%5B%5Cfrac%7Bt%7D%7B2%7D%5Csqrt%7B1%2Bt%5E2%7D%2B%5Cfrac%7B1%7D%7B2%7Dln%28t%2B%5Csqrt%7B1%2Bt%5E2%7D%29%5D%5E%7B1%7D_%7B-2%7D)
Length of curve=![s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B12%7D%5B%5Cfrac%7B12x%2B14%7D%7B2%7D%5Csqrt%7B1%2B%2812x%2B14%29%5E2%7D%2B%5Cfrac%7B1%7D%7B2%7Dln%2812x%2B14%2B%5Csqrt%7B1%2B%2812x%2B14%29%5E2%7D%29%5D%5E%7B1%7D_%7B-2%7D)
Length of curve=
Length of curve=
Length of curve=
The smallest region that can be photographed is
5*7 = 35 square km
so we want to know how long it takes to zoom out to an area of
35 * 5 = 175 square km.
Notice that the length and width increase at the same rate, 2 km/sec, and they start out with a difference of two, so when the width has increased from 5 to x, the length will have increased from 7 to x+2. At the desired coverage area, then,
x(x+2) = 175
x^2 + 2x = 175
x^2 + 2x + 1 = 176 [completing the square]
(x+1)^2 = 176
x+1 = ±4√11
x = -1 ±4√11
Since a negative value for x is meaningless here, x must be
-1 + 4√11 = about 12.27 km
and the time it took to increast to that value was
(12.27 - 5) km / (2 km/sec) = about 3.63 seconds
Answer:
<h2>(-4,-2)</h2>
Step-by-step explanation:
(-4,-5)
move up 3 units
change in y-axis
-5+3=-2
(-4,-2)
Answer:
Non linear. It is a curve not a line.
