Q. 1 - Calculus

Question

1. True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: If $z=f(x, y)$ and $x=u(s, t)$, then $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}$.

(b) True or False: If $z=f(x, y)$ and $x=u\left(t\right)$, then $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}$.

(c) True or False: If $z=f(x, y)$, then $\nabla f(x,y)=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$

(d) True or False: If $z=f(x, y)$ and $\nabla f(1,3)=0$, then the graph of $f$ is differentiable at $(1,3)$.

(e) True or False: If $z=f(x, y)$ and $\mathbf{u}$ is a unit vector, then $D_{\mathrm{u}}f(a,b)=\nabla f(a,b)·\mathbf{u}$.

(f) True or False: If $f(x, y, z)$ is differentiable at $(a, b, c)$ and $\mathbf{u}\in {\mathrm{ℝ}}^3$ is a unit vector, then $D_{u}f(a,b,c)=\nabla f(a,b,c)$.

$\mathbf{u}$.

(g) True or False. If $\backslash (w=f(x, y, z)\backslash )$ is differentiable at $(a, b, c)$. then $-\nabla f(a,b,c)$ points in the direction in which $f$ is decreasing most rapidly at $(a, b, c)$.

(h) True or False: If $z=f(x, y)$ is differentiable at $(-1,0)$. and if $f(-1,0)=4$, then $\nabla f(-1,0)$ is orthogonal to the level curve $f(x, y)=4 at (-1,0)$.

Step-by-Step Solution

Answer

1. The statement $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}$is determined as False
2. The statement $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}$is determined as False
3. The statement $\nabla f(x,y)=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$ is determined as False
4. The statement that states the graph of $f$is differentiable at (1,3) is determined as False
5. The statement $D_{u}f(a,b)=\nabla f(a,b)·\mathbf{u}$ is determined as False
6. The statement $D_{u}f(a,b,c)=\nabla f(a,b,c)\times u$is determined as False
7. The statement $-\nabla f(a,b,c)$ points in the direction in which $f$ is decreasing most rapidly at $(a, b, c)$ is determined as True
8. The statement and $\nabla f(-1,0)$ is orthogonal to the level curve $\$f(x, y)=4\$ at \$(-1,0)\$$is determined as True.

Introduction

The given data is the chain rule and gradient statements

The objective is to determine the statements are true or false

Step 1 : Part (a)

Consider the following statement, "If $\$Z=f(x, y)\$ and \$x=u(s, t)\$$, then $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}$,

then the value of $\frac{\partial z}{\partial s}$ from chain rule formulas is

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}·\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}·\frac{\partial y}{\partial s}.$

Hence, the statement is False

Step 2: Part(b)

Consider the following statement, "If $Z=f(x, y) and x=u\left(t\right)$ then $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}·\frac{\partial x}{\partial t}$ ".

If $Z=f(x, y), x=u\left(t\right)$ and $y=v\left(t\right)$ for all values of $\mathrm{t}$ at which $\mathrm{u}$ and $\mathrm{v}$ are differentiable, and if the function is differentiable at $\left(u\right(t), v(t\left)\right)$, then $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}·\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}·\frac{\partial y}{\partial t}$.

Step 3: Part (c)

Consider the following statement, "If $Z=f(x, y)$ then $\nabla f(x,y)=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$ ".

If $Z=f(x, y)$ be a function of two variables. The gradient of the function is the vector function defined by $\nabla f(x,y)=\mathbf{i}\frac{\partial f}{\partial x}+\mathbf{j}\frac{\partial f}{\partial y}$.

Hence, the statement is False

Step 4: part (d)

Need to determine whether the following statement is true or false The derivative of $\ln \left|x\right|$ is

$\frac{d}{dx}\left(\right|\ln x\left|\right)=\frac{1}{x}$

Thus the given statement is false.

Step 5 : Part (e)

Consider the following statement, "If $Z=f(x, y)$ and $\mathbf{u}$ is a unit vector, then $D_{u}f\left(a_{;}b\right)=\nabla f(a,b)·\mathbf{u}{·}^n$

Let $f(x, y)$ be a function of two variables and $(a, b)$ be a point in the domain of $f(x, y)$ at which $f(x, y)$ is differentiable. Then, for every unit vector $\mathbf{u}\in {\mathrm{ℝ}}^2$,

$D_{u}f(a,b)=\nabla f(a,b)·\mathbf{u}$

Hence, the statement is False.

Step 6: Part(f)

Consider the following statement, "If $f(x, y, z)$ is differentiable at $(a, b, c)$ and $\mathbf{u}\in {\mathrm{ℝ}}^3$ is a unit vector, then $D_{u}f(a,b,c)=\nabla f(a,b,c)·{\mathbf{u}}^n$.

The gradient vectors are orthogonal to level curves theorem proves that the assertion is valid.

Hence, the statement is. True

Step 7: Part (g)

Consider the following statement, "If $\omega =f(x,y,z)$ is differentiable at $(a, b, c)$, then $-\nabla f(a,b,c)$ Points in the direction in which $f(x, y, z)$ is decreasing most rapidly at $(a,b,c{)}^n$.

Let the function of two or three variables be$f$ and let point in the domain be$P$ of $f$ at which $f$ is differentiable.

Then, the gradient of $f$ at $P$ points in the direction in which $f$ increases most rapidly and - the gradient of $f$ at $P$ points in the direction in which $f$ decreases most rapidly.

Hence, the statement is True.

Step 8: Part(h)

Consider the statement, "If $Z=f(x, y)$ is differentiable at, and if $f(-1,0)=4$ then $\nabla f(-1,0)$ is orthogonal to the level curve $f(x, y)=4$ at $(-1,0)$.

The gradient vectors are orthogonal to level curves theorem proves that the assertion is valid.

Hence, the statement is True