Question

id no: section: part i:write true if the statement is correct or false otherwise on the space provided. (1 pt. each) 1. the derivative of a vector - valued function is found by differentiating each of its component functions separately. 2. for a function of two variables, if the limit along two different paths gives different values then the limit exists. 3. a vector - valued function can have scalar outputs. 4. every polynomial function of several variable is continuous everywhere. 5. the binormal vector is found by taking the cross product of the unit tangent and normal vectors. 6. a real - valued function is a function whose values are real numbers. part ii: put the most simplified answer on the space provided (1.5 pts each) 1. the formula of normal vector n(t) is 2. the equation f(x,y)=k represents a of the function f. 3. lim_(x,y)→(1, - 1) (x^(2)-y^(2))/(3x + 3y)= 4. the magnitude of f(t)=(sin t,cos t,2) is part iii: choose the best answer and write your answer on the space provided.(1.5pts each) 1. continuity at a point requires. a. the limit exists and equals the function value c. the function is differentiable b. the limit exists but the function value is different. d. the limit does not exist. 2. the domain and range of f(x,y)=(sqrt(16 - x^(2)-y^(2)))/(x^(2)+y^(2)+1) is a. domain x^(2)+y^(2)≤16 and range 0,4) c. domain x^(2)+y^(2)≤16 and range (0,4 b. domain x^(2)+y^(2)>16 and range (0,4) d. domain all real numbers and range (0,4 3. let f(t)=(t,sin t,t^(2)), then f(t) is a. (1,cos t,2t) b. (0,cos t,2) c. (0, - sin t,2) d(0,cos t,2).

Explanation:

Part I:

1.

The derivative of a vector - valued function $\vec{r}(t)=\langle x(t),y(t),z(t)
angle$ is $\vec{r}'(t)=\langle x'(t),y'(t),z'(t)
angle$, so we differentiate each component function separately. The statement is True.

2.

For a function of two variables $z = f(x,y)$, if the limit along two different paths gives different values, then the limit $\lim_{(x,y)\to(a,b)}f(x,y)$ does not exist. The statement is False.

3.

A vector - valued function has vector outputs, not scalar outputs. The statement is False.

4.

A polynomial function of several variables is continuous everywhere. The statement is True.

5.

The binormal vector $\vec{B}(t)=\vec{T}(t)\times\vec{N}(t)$ where $\vec{T}(t)$ is the unit tangent vector and $\vec{N}(t)$ is the unit normal vector. The statement is True.

6.

A real - valued function $y = f(x)$ (or in multiple variables) has real numbers as its values. The statement is True.

Part II:

1.

The normal vector $N(t)=\frac{T'(t)}{\vert T'(t)\vert}$, where $T(t)=\frac{r'(t)}{\vert r'(t)\vert}$ and $r(t)$ is a vector - valued function.

2.

The equation $f(x,y)=k$ represents a level curve of the function $f$.

3.

We need to simplify the limit $\lim_{(x,y)\to(1, - 1)}\frac{x^{2}-y^{2}}{3x + 3y}=\lim_{(x,y)\to(1, - 1)}\frac{(x - y)(x + y)}{3(x + y)}=\lim_{(x,y)\to(1, - 1)}\frac{x - y}{3}=\frac{1-(-1)}{3}=\frac{2}{3}$

4.

If $\vec{f}(t)=\langle\sin t,\cos t,2
angle$, then $\vert\vec{f}(t)\vert=\sqrt{\sin^{2}t+\cos^{2}t + 4}=\sqrt{1 + 4}=\sqrt{5}$

Part III:

1.

Continuity at a point requires that the limit exists and equals the function value. So the answer is A.

2.

For the function $f(x,y)=\frac{\sqrt{16 - x^{2}-y^{2}}}{x^{2}+y^{2}+1}$, we need $16 - x^{2}-y^{2}\geq0$, so $x^{2}+y^{2}\leq16$. Also, since $x^{2}+y^{2}+1>0$ for all real $x$ and $y$. The range: Let $u = x^{2}+y^{2}$, then $f(x,y)=\frac{\sqrt{16 - u}}{u + 1}$, $0\leq u\leq16$. When $u = 0$, $f(x,y) = \frac{\sqrt{16}}{0 + 1}=4$, when $u = 16$, $f(x,y)=0$. The range is $[0,4]$. The answer is A.

3.

If $f(t)=\langle t,\sin t,t^{2}
angle$, then $f'(t)=\langle1,\cos t,2t
angle$ and $f''(t)=\langle0,-\sin t,2
angle$. The answer is C.

Answer:

Part I:

1. True
2. False
3. False
4. True
5. True
6. True

Part II:

1. $N(t)=\frac{T'(t)}{\vert T'(t)\vert}$
2. level curve
3. $\frac{2}{3}$
4. $\sqrt{5}$

Part III:

1. A. the limit exists and equals the function value
2. A. domain $x^{2}+y^{2}\leq16$ and range $[0,4]$
3. C. $(0,-\sin t,2)$