Exercise 12 solutions

🔹 must know, 🔸 ideally should know, ⛑ challenging, 🔫 very challenging

Exercise from the book

🔹 14.3.13

Find the first partial derivatives of the function \(z = ln(x + t^2)\)

Solution

We see that \(z\) is dependent on two variables, namely \(x\) and \(t\), thus we need to find first derivative of \(z\) with respect to both of these variables as follows:

\[\frac{\partial z}{\partial x} = \frac{1}{x + t^2} \cdot 1 = \frac{1}{x + t^2}\] \[\frac{\partial z}{\partial t} = \frac{1}{x + t^2} \cdot 2t = \frac{2t}{x + t^2}\]

We differentiate with respect to outer function and then multiply it by the inner function. In addition, recall that \(\frac{d}{dx} [\ln{x}] = \frac{1}{x}\).


🔹 14.3.67

If \(f(x, y, z) = xy^2z^3 + \arcsin{(x\sqrt{z})}\), find \(f_{xzy}\).

(Hint: Which order of differentiation is the easiest?)

Solution

Using Clairauts theorem (p. 967 in the book), we know that for example \(f_{xzy} = f_{yzx}\) etc. We can clearly see that we want to avoid differentiating the term \(\arcsin{(x\sqrt{z})}\), thus we should choose the order of differentiation accordingly and start with \(y\) as follows:

\[f_y = 2yxz^3\]

And now, it is easy:

\[\begin{aligned} f_{yx} = 2yz^3 \\ f_{yxz} = 6yz^2 \end{aligned}\]

And because of the Clairauts theorem, we can write \(f_{yxz} = f_{xzy} = 6yz^2\).


🔹 14.4.5

Find the equation of the tangent plane to the given surface at the specified point: \(z = e^{x - y} \text{, } (2, 2, 1)\)

Solution

Following the book (p. 975), we know that equation of a tangent plane to the given surface \(z = f(x, y)\) at the specified point \(P(x_{0}, y_0, z _0)\) can be found as follows:

\[z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\]

Therefore, we start by finding \(f_x\) and \(f_y\):

\[f_x(x, y) = e^{x - y}\] \[f_y(x, y) = -e^{x - y}\]

And then we evaluate both derivatives at the point \((x_0, y_0) = (2, 2)\):

\[f_x(2, 2) = 1\] \[f_y(2, 2) = -1\]

Finally, we can plug the resuls to the general equation above as follows and obtain the equation for our tangent plane:

\[\begin{aligned} z - 1 = (x - 2) - (y - 2) \\ z = x - y + 1 \end{aligned}\]


⛑ 14.4.52

Suppose you need to know an equation of the tangent plane to a surface \(S\) at the point \(P(2, 1, 3)\). You do not know an equation for \(S\) but you know that the curves

\[\begin{aligned} r_1 = \langle 2 + 3t, 1 - t^2, 3 - 4t + t ^2 \rangle \\ r_2 = \langle 1 + u^2, 2u^3 - 1, 2u + 1 \rangle \end{aligned}\]

both line on \(S\). Find an equation of the tangent plane at \(P\).

Solution

If you look at page 974 in the book and figure 1, you can see that there we have quite similar case. You are given two curves which intersect at some point \(P\). Tangent lines to these curves are actully part of the plane which is tangent to the surface \(S\) at the point \(P\). If you read the paragraph Tangent planes on the same page, you should spot one important phrase:

... if C is any other curve that lies on the surface S and passes through P, then its tangent line at P also lies in tangent plane. Therefore you can think of the tangent plane to S at P as consisting of all posible tangent lines at P to curves that lie on S and pass though P.

Now, you may ask, how is this relevant to our problem? Well, if we know that the two curves we are given are going through the point \(P\), then we know that their corresponding tangent vectors at this point \(P\) lie on the tangent plane for which we want to find the equation. I will explain later why are these tangent vectors useful in helping us finding the equation of the tangent plane.

Therefore, let’s start with finding whether the two curves go even through the point \(P\). In practice, this means, that for curve \(r_1\), we want to know if there exists some \(t\) such that \(r_1(t) = (2, 1, 3)\). And similar logic applies to the curve \(r_2\). Therefore, to find such \(t\) for curve \(r_1\), we can write:

\[\begin{aligned} 2 + 3t = 2 \\ 1 - t^2 = 1 \\ 3 - 4t + t^2 = 3 \end{aligned}\]

From the first equation, I can see that \(t = 0\). If I plug \(t = 0\) to the other equations, the equality still holds, thus for \(r_1\), we know that \(t = 0\). Using the same procedure for \(r_2\), we arrive at \(u = 1\). We continue with computing first derivatives of the two curves:

\[r_1'(t) = \langle 3, -2t, -4 + 2t \rangle\] \[r_2'(u) = \langle 2u, 6u^2, 2 \rangle\]

So, let me summarize what we have computed so far:

  • We know that both curves \(r_1\) and \(r_2\) go through the point \(P\), they do so for \(t = 0\) and \(u = 1\) respectively

  • We computed first derivatives of the two curves. You should know what these represent, if not, refer to the previous lecture.

As a next step, let’s compute our tangent vectors, i.e., vectors that are tangent to corresponding curve at point \(P\). This can be done as follows:

\[r_1'(0) = \langle 3, 0, -4 \rangle\] \[r_2'(1) = \langle 2, 6, 2 \rangle\]

Now, we are at the step where it is important to think how we can use these two vectors to obtain our tangent plane.

During linear algebra, you were introduced to the notion of dot product which yields scalar. There is also cross product which we did not talk about. Cross product of two vectors yields another vector, this vector happens to be orthogonal to the given two vectors which you use in cross product. The way you can imagine it is that you have two independent vectors in 2D plane and if you take their cross product, you obtain a third vector which is orthogonal to them, which means that this vector spans the third dimension. If we know this orthogonal or normal vector, then we can use the equation for plane defined on page 868 in the book: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\) where \(n = \langle a, b, c \rangle\). (\(n\) stands for normal vector). I will use python to compute the cross product for us:

import numpy as np

r1 = np.array([3, 0, -4])
r2 = np.array([2, 6, 2])

np.cross(r1, r2)

The code returns the following vector \([ 24, -14, 18]\).

Finally, we can conclude that the tangent plane at the point \(P\) is defined as:

\[24(x - 2) - 14(y - 1) + 18(z - 3) = 0\]


🔸 14.5.39

The temperature at a point \((x, y)\) is \(T(x, y)\), measured in degrees Celsius. A bug crawls so that its position after \(t\) seconds is given by \(x = \sqrt{1 + t}\) and \(y = 2 + \frac{1}{3}t\) where \(x\) and \(y\) are measured in \(cm\). The temperature functions satisifies \(T_x(2, 3) = 4\) and \(T_y(2, 3) = 3\). How fast is the temperature rising on the bug’s path after 3 seconds.

Solution

First, we can see that \(T\) is dependent on \(x, y\) and \(x, y\) are dependent on \(t\), thus by transitive property, we can see that \(T\) depends on \(t\). Therefore, we are interested in rate of change of \(T\) when \(t = 3\), in other words, we want to know \(\frac{d}{d t} T(x(3), y(3))\).

We should notice that if \(t = 3\), then we know that \(x = 2\) and \(y = 3\). In addition, we know that:

\[\frac{\partial}{\partial x} T(2, 3) = 4\] \[\frac{\partial}{\partial y} T(2, 3) = 3\]

We should recall Chain Rule - case 1 from the book, i.e.:

\[\frac{d}{d t} T = \frac{\partial T}{\partial x}\frac{d x}{d t} + \frac{\partial T}{\partial y}\frac{d y}{d t}\]

Therefore, we can write:

\[\frac{d}{d t} T(x(3), y(3)) = 4\frac{d}{d t}x(3) + 3\frac{d}{d t}y(3)\]

Finally, we need to figure out the derivatives with respect to \(t\):

\[\frac{d x}{d t} = \frac{1}{2}(1 + t)^{-\frac{1}{2}}\] \[\frac{d y}{d t} = \frac{1}{3}\]

Therefore, the rate of change of temparature after 3 seconds is:

\[\frac{d}{d t} T(2, 3) = 4[\frac{1}{2}(1 + 3)^{-\frac{1}{2}}] + 3\frac{1}{3} = 4 \frac{1}{2}\frac{1}{\sqrt{4}} + 1 = 4\frac{1}{4} + 1 = 2\]


Exercise from the Rasmus’ weekly note

⛑ Exercise 0.1

a) Consider the function \(f(x, y) = xy\), and suppose \(g, h : R → R\) are two functions, that we do not yet know what are. Use the chain rule to compute the derivative of the function \((f \circ \langle g, h \rangle)(t) = f(g(t), h(t))\). This should give you a well known rule. Which one is it?

b) Consider the functions \(f(x, y) = x^y\) and \(r(t) = (t, t)\). Compute \((f \circ r)'(t)\) using the chain rule. Could you have computed this derivative directly?

(Hint: \((a^x)' = ln(a)a^x\))

Solution

a)

We can use the formula for Chain rule - case 1 again and write:

\[\frac{d}{d t} f = \frac{\partial f}{\partial x}\frac{d x}{d t} + \frac{\partial f}{\partial y}\frac{d y}{d t}\]

Now, we know that:

\[\frac{\partial f}{\partial x} = y = h(t)\] \[\frac{\partial f}{\partial y} = x = g(t)\] \[\frac{d x}{d t} = g'(t)\] \[\frac{d y}{d t} = h'(t)\]

So, if we plug that to our initial formula, we obtain:

\[\frac{d}{d t} f = h(t)g'(t) + h'(t)g(t)\]

And this is indeed Product rule which was introduced to you in the lecture 10.

b)

Again, we can use the formula for Chain rule - case 1 and write:

\[\frac{d}{d t} f = \frac{\partial f}{\partial x}\frac{d x}{d t} + \frac{\partial f}{\partial y}\frac{d y}{d t}\]

Then, we know that:

\[\frac{\partial f}{\partial x} = yx^{y - 1} = tt^{t -1} = t^t\] \[\frac{\partial f}{\partial y} = ln(x)x^y = ln(t)t^t\] \[\frac{d x}{d t} = 1\] \[\frac{d y}{d t} = 1\]

And therefore:

\[\frac{d}{d t} f = t^t + ln(t)t^t = t^t(1 + ln(t))\]

Clearly, if we wanted to compute it directly, this would mean differentiating \(t^t\) which would be impossible since we would not know whether to treat the function as exponential or power function given we have \(t\) both as base as well as power.

🔍 About the page

  • Last updated: 23/11/2022
  • Unless othewise stated, exercises come from the book: James Stewart, Daniel K. Clegg and Saleem Watson: Calculus: Early Transcendentals, Metric Edition, 9th edition
  • Purpose: This page was created as part of a preparation for my teaching assistant session in the course Linear algebra and optimization managed by Rasmus Ejlers Møgelberg. Please note that the solutions on this page are my own, thus it should not be considered as official solutions provided as part of the course.