Answer:
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
