Calculus 3 Practice Exam — Free

Let $g(x,y,z) = x^2 + y^2 + z^2 - 49$. At a constrained extremum the gradients are parallel, $\nabla f = \lambda \nabla g$, which together with the constraint gives four equations in four unknowns:

None of the components of $\nabla f$ is zero, so $\lambda \ne 0$ and each equation can be solved for a coordinate: $x = \dfrac{1}{\lambda}$, $y = \dfrac{3}{2\lambda}$, $z = \dfrac{3}{\lambda}$. Substitute into the constraint to pin down $\lambda$:

$\lambda = \tfrac12$ gives the point $(2, 3, 6)$ and $\lambda = -\tfrac12$ gives $(-2, -3, -6)$. The sphere is closed and bounded and $f$ is continuous, so the maximum and minimum exist and must be among these candidates:

As a cross-check, the Cauchy–Schwarz inequality gives $|2x + 3y + 6z| \le \sqrt{2^2+3^2+6^2}\,\sqrt{x^2+y^2+z^2} = 7 \cdot 7 = 49$, with equality exactly when $(x,y,z)$ is parallel to $(2,3,6)$.

The tempting wrong answer is 42, from putting the whole "budget" into the variable with the largest coefficient, $(0, 0, 7)$; that point satisfies the constraint, but the gradients are not parallel there, so it is not an extremum. The answer 77 multiplies the sum of the coefficients by the radius, which has no geometric meaning here, and 14 is the value at $(7, 0, 0)$.

The maximum value is 49, attained at $(2, 3, 6)$ (option C).

The directional derivative in the direction of a unit vector $\mathbf{u}$ is $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$, so two things are needed: the gradient at the point and a unit direction vector.

Gradient. Using the product and chain rules,

At $(2, 0)$: $f_x = 0 + 0 = 0$ and $f_y = 4 + 2e^{0} = 6$, so $\nabla f(2,0) = \langle 0, 6\rangle$.

Direction. The displacement from $(2, 0)$ to $(5, 4)$ is $\langle 3, 4\rangle$, which has length $\sqrt{9 + 16} = 5$. The direction must be a unit vector (otherwise the answer would scale with the length of the arrow), so

Dot product:

Since the gradient points straight up the $y$-direction with magnitude 6, moving in a direction whose vertical component is $\tfrac45$ gives $\tfrac45$ of the maximum rate of change, which is a sensible result.

$D_{\mathbf{u}} f(2, 0) = \dfrac{24}{5} = 4.8$

Along the straight segment the integral would require $\displaystyle\int_{-3}^{3} e^{x^2}\,dx$, which has no elementary antiderivative, so a direct computation is impractical; Green's theorem converts the line integral into a double integral over the region $D$ enclosed by $C$. With $P = e^{x^2} - y^2$ and $Q = x^2 + \cos y$,

The orientation is counterclockwise (region on the left), as Green's theorem requires, so no sign change is needed. The awkward terms $e^{x^2}$ and $\cos y$ disappear because each depends only on its own variable.

The region is the upper half of a disk of radius 3, so polar coordinates are natural: $x = r\cos\theta$, $y = r\sin\theta$, $0\le r\le 3$, $0\le\theta\le\pi$, $dA = r\,dr\,d\theta$.

By symmetry, $\iint_D 2x\,dA = 0$: the half-disk is symmetric about the $y$-axis and $2x$ is odd in $x$. (Explicitly, $\int_0^{\pi}\cos\theta\,d\theta = 0$.) For the other term,

Therefore the line integral equals $0+36 = 36$.

$\displaystyle\oint_C \left(e^{x^2} - y^2\right)dx + \left(x^2 + \cos y\right)dy = 36$