Compute the directional derivatives of the following functions along unit vectors at the indicated points in directions parallel to the given vector.
(a) f(x, y) = xy, (x0, y0) = (e, e), d = 20i + 21j
(b) f(x, y, z) = ex + yz, (x0, y0, z0) = (1, 1, 1), d = (3, −5, 5

Answer of part(a) :41*e/29 ≈ 3.84308810230417

Answer of part(b):(3*((59)^1/2)*e)/59≈1.06167045295814

Step-by-step explanation:

Solution of Part(a)

First we find the directional derivative of  

f(x,y) = xy at (x0,y0)=(e,e) in the direction of the vector d⃗ =(20,21)

To find the gradient of a function (which is a vector), differentiate the function with respect to each variable.

∇f=(∂f/∂x,∂f/∂y)

df/dx=d(xy)/dx  

here we Apply the constant multiple rule ddx(c⋅f(x))=c⋅ddx(f(x))ddx(c⋅f(x))=c⋅ddx(f(x)) with c=yc=y and f(x)=x

so we get

df/dx = y

similarly

d(xy)/dy = x

Finally, plug in the point:

∇f(e,e)=(e,e)

∇(xy)(x0,y0)=(y,x)

∇(xy)|(x0,y0)=(e,e)=(e,e)

Find the length of the vector: |d⃗ |=((20)^2+(21)^2)^1/2 = 29

To normalize the vector, divide each component by length:

d⃗ d→ becomes (20/29,21/29)

Finally, the directional derivative is the dot product of the gradient and the normalized vector:

D(xy)d⃗ (e,e)   =(e,e)⋅(20/29,21/29)

                    = 41*e/29 ≈ 3.84308810230417

so this is our  answer.

Solution of Part(b)

We have to find the directional derivative of the given function  

f(x,y,z) = yz+e^x at (x0,y0,z0)=(1,1,1) in the direction of the vector

d⃗ =(3,−5,5)  

Find the gradient of the function and evaluate it at the given point:  

To find the gradient of a function (which is a vector), differentiate the function with respect to each variable.  

∇f= (∂f/∂x,∂f/∂y,∂f/∂z)  

∂f/∂x = ∂( yz+e^x )/∂x  

The derivative of a sum/difference is the sum/difference of derivatives:  

∂f/∂x = ∂( yz)/∂x +∂(e^x )/∂x since the derivative of y and z with respect of x is zero.  

∂f/∂x = e^x The derivative of exponential is d(e^x)/dx=e^x.  

Similarly  

d(yz+e^x)/dy = z and d(yz+e^x)/dz = y  

∇(yz+e^x)(x0,y0,z0)=(e^x,z,y)  

Finally plug in the point we get the result  

∇(yz+e^x)|(x0,y0,z0)=(1,1,1)=(e,1,1)  

∇(yz+ex)|(x0,y0,z0)=(1,1,1)=(e,1,1)  

Now we Find the length of the vector: |d⃗ |=((3)^2+(−5)^2+(5)^2)^1/2 =(59)^1/2  

To normalize the vector, divide each component by length:  

d = ((3*((59)^1/2))/59 ,(−5*(59)^1/2)/59,(5*(59)^1/2)/59)  

Finally, the directional derivative is the dot product of the gradient and the normalized vector:  

D(yz+e^x)d(1,1,1) = (e,1,1). ((3*((59)^1/2))/59 ,(−5*(59)^1/2)/59,(5*(59)^1/2)/59)  

D(yz+e^x)d(1,1,1) = (3*((59)^1/2)*e)/59≈1.06167045295814  

So this is our answer.

In the "*" shows multipalication

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )