Calculus Problem Set: Derivatives, Integrals, and Series
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3. Find all first and second order partial derivatives of $$f(x, y) = x^2 + 6y^5 - 10xy.$$
4. Construct the Taylor series expansion of
(a) $f(x) = \sin x$ about $x = \frac{\pi}{2}$.
(b) $f(x) = \ln x$ about $x = 1$.
(c) $f(x) = \frac{1}{x^2}$ about $x = 2$.
5. Find the length of the arc of the curve $y = (x - 1)^{\frac{3}{2}}$ from $x = 1$ to $x = 5$.
6. In the Taylor series expansion of $f(x) = x^4$ centered at $a = 1$, what is the coefficient of $(x - 1)^3$.
7. Consider the curve parameterized by $$x = \frac{1}{3}t^3 + 3t^2 + 2, \quad y = t^3 - t^2,$$
(a) Find an equation for the line tangent to the curve when $t = 1$.
(b) Compute $\frac{d^2y}{dx^2}$ at $t = 1$.
8. Use the $t$-method method to evaluate the integral: $\int \frac{d\theta}{1 - \cos \theta}$
9. The velocity at time $t$ of a particle moving along a straight line is given by $$v = t^3 - 4t^2 + 4t.$$
Find the distance the particle will have covered between $t = 2$ and $t = 5$.
10. Calculate the area of the region bounded by the graphs
(a) $f(x) = x^2 + 1$ and $g(x) = 3 - x^2$.
(b) $y = x - 1$ and $y = 0$.
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Hi Marion! Let's solve this calculus problem where we find the distance covered by a particle between two points in time given its velocity function.
Problem 9: Distance Calculation
The velocity function is given as v of t equals t cubed minus four t squared plus four t. We want to find the distance between time equals two and time equals five.
Recall that distance is the integral of the speed, which is the absolute value of velocity. Let's factor the expression to see where the velocity might be negative.
Notice that the term in the parentheses is a perfect square. So, v of t simplifies to t multiplied by the quantity t minus two squared.
Since t is between two and five, t is positive. Also, any number squared is non-negative. Therefore, the velocity is greater than or equal to zero on our entire interval.
Because the velocity is never negative, the total distance is simply the definite integral of the velocity function from two to five.
Now, we perform the integration term by term using the power rule.
Integrating the Velocity
The integral of t cubed is one fourth t to the power of four. The integral of four t squared is four thirds t cubed. And the integral of four t is two t squared.
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