Hi, can someone help me solve these 5 questions please? Thank you! 1) 2) 3) 4) 5

Question

Hi, can someone help me solve these 5 questions please? Thank you!

1)

2)

3)

4)

5)

Given that Calculate f'(-1). f(x) = x 9h(x) h(-1)=4 h'(-1) = 7 Use implicit differentiation to find the slope of the tangent line to the curve at the point (1. -9 / -71). y / x + 8y =x2 + 8 The altitude of a triangle is increasing at a rate of 1.000 centimeters/minute while the area of the triangle is increasing at a rate of 3.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.000 centimeters and the area is 87.000 square centimeters? A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.1 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 15 cm. At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 23 knots and ship B is sailing north at 20 knots. How fast (in knots) is the distance between the ships changing at 6 PM?

Explanation / Answer

1. f(x)=x^9 h(x)

=> f'(x) = 9x^8 h'(x) => f'(-1) = 9 x 1 x h'(-1) = 9 x7 =63

2) y =(x^2 +8 )(x+8y)

=> dy/dx = 2x(x+8y) + (x^2+8)(1+8 dy/dx)

putting x=1 and y=-9/71 in the above equation, we get :

dy/dx = 637/71 + 72 dy/dx => dy/dx =637/5041 = 0.12636

3) Area = 0.5 x b x h => dA/dt = 0.5 ( b x dh/dt + h x db/dt)

given , dA/dt = 3 dh/dt=1 h=9 A =87

thus, b=58/3

6= 58/3 x 1 + 9 x db/dt => db/dt = -40/27 =-1.4814 cm/s

4) V= 1/6 x pi x D^3

=> dV/dt = pi/2 x D^2 x dD/dt = (45 x pi/4) cm^3/min

5) The distance between them is given by :

r^2 = x^2+ y^2

=> 2r dr/dt = 2x dx/dt + 2y dy/dt . . . . . . . . (1)

(x is the distance dtavelled by A and y is the distance travelled by B)

at 6pm , x=20 + 23x6 = 158 nautical miles

y = 20 x 6 =120 nautical miles r= 198.4 nautical miles

dx/dt = 23 knots dy/dt=20 knots

putting these in the above equation (1) , we get :

2 x 198.4 x dr/dt = 2x158x23 + 2x120x20

thus, dr/dt =30.4133 knots

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