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Natalka [10]
3 years ago
11

In a concave mirror, if an object is located behind the center of the curvature, What is the best description of the Image forme

d?
Physics
1 answer:
kipiarov [429]3 years ago
5 0
 be either real or virtual.
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(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat
vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i)  the gradient of f =  ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is  ñ∙∇F =  4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+  ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+  (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have  

∇f(x, y) = -1·i -3·j  that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.  

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z)  = ∇F(x, y, z) = = ∂f/∂x i+  ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore  

ñ = ⅟4(2·i +3·j -√3·k)  

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)  

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

8 0
3 years ago
What was Marie's speed from point A to point B, in miles/minute?
sveticcg [70]

Answer:

40m

Explanation:

40m

6 0
2 years ago
An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?
spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

6 0
3 years ago
When does the most significant drop in physical activity typically occur in people's lives?
Eduardwww [97]

Answer:

Later high school years and freshman year of college

Explanation:

The transition from high school to college is an important developmental milestone that holds the potential for personal growth and behavioral change.  A cohort of 2,025 students was recruited during the summer before they matriculated into college and completed Internet-based surveys about their participation in a variety of behavioral risks during the last three months of high school and throughout the first year of college. Alcohol use, marijuana use, and sex with multiple partners increased during the transition from high school to college, whereas driving after drinking, aggression, and property crimes decreased. Those from rural high schools and those who elected to live in private dormitories in college were at highest risk for heavy drinking and driving after drinking.

4 0
3 years ago
When in orbit, astronauts experience weightlessness what is this caused by?
erica [24]

no gravity or dark matter

4 0
3 years ago
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