math 8: calculus in one and several ariablesv spring 2018 ...m8s18/math8_hw7_s18_sol.pdfmath 8:...

Math 8: Calculus in one and several variablesSpring 2018 - Homework 7

Return date: Wednesday 05/16/18

keywords: tangent planes, chain rule in several variables

exercise 1. (3 points) For the following functions �nd the indicated partial derivatives.

a) f(x, y) = x4y − x3y3. Find fxxx and fyxy.

Solution: When calculating fx = ∂f(x,y)∂x we treat y as a constant. We get

fx =∂

∂xx4y − x3y3 = 4x3y − 3x2y3.

In a similar fashion we get for fxx = ∂fx(x,y)∂x and fxxx = ∂fxx(x,y)

∂x :

fxx =∂

∂x4x3y − 3x2y3 = 12x2y − 6xy3 and fxxx =

∂

∂x12x2y − 6xy3 = 24xy − 6y3.

For fyxy we get

fy =∂

∂yx4y − x3y3 = x4 − 3x3y2,

fyx =∂

∂xx4 − 3x3y2 = 4x3 − 9x2y2 and fyxy =

∂

∂y4x3 − 9x2y2 = −18x2y.

b) f(x, y, z) = y · ex3+y2+z3 . Find fyz.

Solution: When calculating fy = ∂f(x,y,z)∂y we now treat x and z as constants. We get

fy =∂

∂y(y · ex2+y2+z3) = ex

3+y2+z3 + 2y2ex3+y2+z3 .

Then fyz =∂fy(x,y,z)

∂z :

fyz =∂

∂z(ex

3+y2+z3 + 2y2ex3+y2+z3) = 3z2ex

3+y2+z3 + 6y2z2ex3+y2+z3 .

exercise 2. (3 points) Find an equation for the tangent plane to the graph of the givenfunction at the given point.Solution: In general the equation for the tangent plane TP at the point

P = (x0, y0, z0) = (x0, y0, f(x0, y0)) is

TP : z = z0 + fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0).



a) f(x, y) = x2 + 3y2 at the point P = (1, 1, 4). Sketch the function and the plane.Solution: We �rst have to calculate the partial derivatives fx and fy. We get:

fx(x, y) = 2x , fx(1, 1) = 2 and fy(x, y) = 6y , fy(1, 1) = 6.

Then from the formula above we get the equation for the tangent plane at P = (1, 1, 4):

TP : z = 4 + 2(x− 1) + 6(y − 1) or 2x+ 6y − z = 4.

To sketch the function we can look at the level sets and the cross sections.For the level sets Lf (k) of f we get: k = z = f(x, y) = x2 + 3y2 ≥ 0 ⇔ k = x2 + 3y2.Hence

Lf (k) = {(x, y) ∈ R2, such that k = x2 = 3y2} (k ≥ 0).

The level sets are ellipses in the xy-plane. A level curve Lf (k) can be parametrized by

Lf (k) : (x, y) =

(√k · cos(t),

√k

3· sin(t)

)where t ∈ [0, 2π].

Then x2 + 3y2 = (√k cos(t))2 + 3(

√k3 · sin(t))2 = k.

Additionally we can look at the sections of the function f . For y = 0 and x = 0 we get:

y = 0 : z = f(x, 0) = x2 + 02 = x2 (xz − plane)

x = 0 : z = f(0, y) = 0 + 3y2 = 3y2 (yz − plane)

Hence in the cross sections we get z = x2 and z = 3y2. Using this information togetherwith the information about the level sets, we can sketch the graph Γ(f) of the function f :

Γ(f)



b) f(x, y) = ex+2y at the point P = (−2, 1, 1).Solution: We �rst have to calculate the partial derivatives fx and fy. We get:

fx(x, y) = ex+2y , fx(−2, 1) = e0 = 1 and fy(x, y) = 2 · ex+2y , fy(−2, 1) = 2 · e0 = 2.

Then from the formula above we get the equation for the tangent plane at P = (−2, 1, 1):

TP : z = 1 + 1 · (x+ 2) + 2 · (y − 1) or x+ 2y − z = −1.

exercise 3. (3 points) Find the approximation of f (i.e., the linearization) of the functionat the given point.

a) f(x, y) =√xy + 2 at (x, y) = (4, 1).

Solution: The equation for the tangent plane TP at the point P = (x0, y0, f(x0, y0)) is


This is a function TP (x, y) = L(x, y), the linear approximation of f near (x, y) = (4, 1).We get:

fx(x, y) =y

2√xy + 2

, fx(4, 1) =1

2√

4 · 1 + 2=

1

2√

6and fy(x, y) =

x

2√xy + 2

, fy(4, 1) =2√6.

With the help of these values and f(4, 1) =√

6 we get the approximation of f near(x, y) = (4, 1) from the equation of the tangent plane TP (x, y) = L(x, y):

L(x, y) = z =√

6 +1

2√

6· (x− 4) +

2√6· (y − 1) or L(x, y) =

1

2√

6x+

2√6y + 3

√6.

b) f(x, y) = x+2y−2 at (x, y) = (0, 0).

Solution: The equation for the tangent plane TP at the point P = (x0, y0, f(x0, y0)) is


This is a function TP (x, y) = L(x, y), the linear approximation of f near (x, y) = (4, 1).We get:

fx(x, y) =1

y − 2, fx(0, 0) = −1

2and fy(x, y) = − (x+ 2)

(y − 2)2, fy(0, 0) = −1

2.

With the help of these values and f(0, 0) = −1 we get the approximation of f near (x, y) =(0, 0) from the equation of the tangent plane TP (x, y) = L(x, y):

L(x, y) = z = −1 +−1

2· (x− 0) +

−1

2· (y − 0) or L(x, y) = −x

2− y

2− 1.



exercise 4. (4 points) Suppose you need to know the tangent plane of a surface S containingthe point P = (2, 1, 3). You do not know the equation for S, but you know that the curves

r1(t) = 〈2 + 3t, 1− t2, 3− 4t+ t2〉 and r2(s) = 〈1 + s2, 2s3 − 1, 2s+ 1〉

both lie in S and pass through P .Find an equation of the tangent plane of S passing through P .Solution: As the curves lie in S their tangent vectors at P lie in the tangent plane TPof S at P . Hence at P the corresponding tangent vectors are perpendicular to the normal vectornP of TP which is given by the equation

TP : 〈(x, y, z)− P 〉 • nP = 0.

We �rst have to �nd these tangent vectors. To this end we have to �nd the values of t and swhere the curves pass P . We get:

r1(t) = 〈2 + 3t, 1− t2, 3− 4t+ t2〉 = 〈2, 1, 3〉 ⇒ 2 + 3t = 2⇒ t = 0 and

r2(s) = 〈1 + s2, 2s3 − 1, 2s+ 1〉 = 〈2, 1, 3〉 ⇒ 2s+ 1 = 3⇒ s = 1.

Then we �nd have to �nd the tangent vectors of r1(t) and r2(s) at P . We have

r′1(t) = 〈3,−2t,−4 + 2t〉 ⇒ r′1(0) = 〈3, 0,−4〉 and r′2(s) = 〈2s, 6s2, 2〉 ⇒ r′2(1) = 〈2, 6, 2〉.

As r′1(0) and r′2(1) are perpendicular to a normal vector nP of TP we can set

nP = r′1(0)× r′2(1) = 〈3, 0,−4〉 × 〈2, 6, 2〉 = 〈24,−14, 18〉 = 2 · 〈12,−7, 9〉.

To simplify our calculations we take the normal vector nP = 〈12,−7, 9〉. Hence with the equationfor the tangent plane we get

TP : 〈(x, y, z)− (2, 1, 3)〉 • 〈12,−7, 9〉 = 0 or 12x− 7y + 9z = 44.

exercise 5. (4 points) Use the chain rule to �nd∂z

∂sand

∂z

∂twhere

z = arctan(x2 + y2) and x = s · ln(t), y = t · es.

Solution: By the chain rule we have:

∂z

∂s=∂z

∂x· ∂x∂s

+∂z

∂y· ∂y∂s

and∂z

∂t=∂z

∂x· ∂x∂t

+∂z

∂y· ∂y∂t.



Step 1: We �rst calculate ∂z∂x and ∂z

∂y .

∂z

∂x=

∂

∂xarctan(x2 + y2) =

2x

(x2 + y2)2 + 1and

∂z

∂y=

2y

(x2 + y2)2 + 1.

Step 2: Then we calculate ∂x∂s and ∂y

∂s and ∂x∂t and ∂y

∂t .

I.)∂x

∂s=

∂

∂ss · ln(t) = ln(t) and

∂y

∂s=

∂

∂st · es = tes.

II.)∂x

∂t=

∂

∂ts · ln(t) =

s

tand

∂y

∂t=

∂

∂tt · es = es.

Step 3: Then we substitute x = s ln(t) and y = tes into ∂z(x,y)∂x and ∂z(x,y)

∂y :

∂z(x, y)

∂x=

2s ln(t)

(s2 ln(t)2 + t2e2s)2 + 1and

∂z(x, y)

∂y=

2tes

(s2 ln(t)2 + t2 · e2s)2 + 1.

Step 4: Plugging in the results from Step 2 and Step 3 into the formula for the chain rule, weget:

∂z

∂s=

2s ln(t) · ln(t)

(s2 ln(t)2 + t2e2s)2 + 1+

2tes · tes

(s2 ln(t)2 + t2 · e2s)2 + 1=

2s ln2(t) + 2t2e2s

(s2 ln(t)2 + t2e2s)2 + 1,

∂z

∂t=

2s ln(t) · st(s2 ln(t)2 + t2e2s)2 + 1

+2tes · es

(s2 ln(t)2 + t2 · e2s)2 + 1=

2s2 ln(t)t + 2te2s

(s2 ln(t)2 + t2e2s)2 + 1.



exercise 6. (3 points) The temperature at a point (x, y) in the plane is givenby the function T (x, y). A bug crawls along a path c, such that its position after time t inseconds is given by

c : r(t) = 〈√

3− t, 3t〉, where t ∈ [0, 10].

From measurements you know that the temperature function satis�es

Tx(√

2, 3) = 2 and Ty(√

2, 3) = 3.

How fast is the temperature rising on the bugs path after 1 seconds?Solution: We know that the change of the temperature along the path r(t) = 〈x(t), y(t)〉 attime t = t0 is

∂T (r(t))

∂t

∣∣∣t=t0

=∂T (x(t), y(t))

∂t

∣∣∣t=t0

= Tx(x(t0), y(t0)) · x′(t0) + Ty(x(t0), y(t0)) · y′(t0).

Here the last equation follows from the chain rule.We know furthermore that

r(t) = 〈√

3− t, 3t〉 = 〈√

2, 3〉 ⇒ 3t = 3⇒ t = 1.

Hence the bug reaches the point (√

2, 3) after one second. With

r′(t) = 〈x′(t), y′(t)〉 =

⟨−1

2√

3− t, 3

⟩, we have r′(1) = 〈x′(1), y′(1)〉 =

⟨− 1

2√

2, 3

⟩.

Using the chain rule we get

∂T (r(t))

∂t

∣∣∣t=1

= Tx(√

2, 3) · x′(1) + Ty(√

2, 3) · y′(1) = − 1√2

+ 9 = 8.293.

Therefore the temperature changes by 8.293◦ per second when the bug reaches the point (√

2, 3).

math 8: calculus in one and several ariablesv spring 2018 ...m8s18/math8_hw7_s18_sol.pdfmath 8:...

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