Question

Find the derivative with respect to \(x\) of the function y represented parametrically (i) \(\mathrm{x}=\int_{0}^{\mathrm{t}} \sin \mathrm{t} \mathrm{dt}, \mathrm{y}=\int_{0}^{\mathrm{t}} \cos \mathrm{t} \mathrm{dt}\); (ii) \(\mathrm{x}=\int_{1}^{t^{2}} \mathrm{t} \ln \mathrm{t} \mathrm{dt}, \mathrm{y}=\int_{t^{2}}^{1} \mathrm{t}^{2} \ln \mathrm{t} \mathrm{dt}\).

Short answer

Question: Find the derivatives of y with respect to x for the following cases: (i) \(x = \int_{0}^{t} \sin t dt\) and \(y = \int_{0}^{t} \cos t dt\) (ii) \(x = \int_{1}^{t^2} t \ln t dt\) and \(y = \int_{t^2}^{1} t^2 \ln t dt\) Answer: (i) \(\frac{dy}{dx} = \frac{\cos t}{\sin t}\) (ii) \(\frac{dy}{dx} = -\frac{t^2 \ln t}{t^2 \ln t - \int_{1}^{t^2} \ln t dt}\)

Step-by-step solution

Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

We know that \(x = \int_{0}^{t} \sin t dt\) and \(y = \int_{0}^{t} \cos t dt\). We can find their derivatives by differentiating both sides with respect to t: \(\frac{dx}{dt} = \frac{d}{dt} (∫_{0}^{t} \sin t dt) = \sin t\) \(\frac{dy}{dt} = \frac{d}{dt} (∫_{0}^{t} \cos t dt) = \cos t\)

Calculate \(\frac{dt}{dx}\)

Instead of calculating \(\frac{dy}{dx}\) directly, we will calculate its reciprocal, \(\frac{dt}{dx}\). This is because the expressions for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are available and much simpler to compute: \(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{\sin t}\)

Calculate \(\frac{dy}{dx}\)

Now, we multiply \(\frac{dy}{dt}\) by \(\frac{dt}{dx}\) to get \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \cos t \cdot \frac{1}{\sin t} = \frac{\cos t}{\sin t}\) This is the derivative of y with respect to x for Case (i). #Case (ii):#

Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

Now we know that \(x = \int_{1}^{t^2} t \ln t dt\) and \(y = \int_{t^2}^{1} t^2 \ln t dt\). We can find their derivatives as before: \(\frac{dx}{dt} = \frac{d}{dt} (∫_{1}^{t^2} t \ln t dt) = 2t( t^2 \ln t - \int_{1}^{t^2} \ln t dt)\) \(\frac{dy}{dt} = \frac{d}{dt} (∫_{t^2}^{1} t^2 \ln t dt) = -2t^3 \ln t\)

Calculate \(\frac{dt}{dx}\)

As before, we will calculate the reciprocal of \(\frac{dy}{dx}\): \(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2t( t^2 \ln t - \int_{1}^{t^2} \ln t dt)}\)

Calculate \(\frac{dy}{dx}\)

Now, we multiply \(\frac{dy}{dt}\) by \(\frac{dt}{dx}\) to get \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = (-2t^3 \ln t) \cdot \frac{1}{2t(t^2 \ln t - \int_{1}^{t^2} \ln t dt)} = -\frac{t^2 \ln t}{t^2 \ln t - \int_{1}^{t^2} \ln t dt}\) This is the derivative of y with respect to x for Case (ii).

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulations of quantities and the areas under and between curves.
When we calculate the integral of a function, we are finding the area under the curve of that function on a graph. There are two main types of integrals: definite and indefinite. Definite integrals have limits of integration and result in a real number representing the area. Indefinite integrals, on the other hand, represent a family of functions and include a constant of integration.

Understanding integrals is vital because they have applications in various fields including physics, engineering, and economics. They help us solve for quantities like distance, area, volume, and other concepts that arise from accumulation.

Derivatives of Parametric Equations

Derivatives of parametric equations involve calculating the rate of change of two variables that are both defined in terms of a third variable, called the parameter. In the exercise provided, for example, both x and y are functions of the parameter t.
When dealing with parametric equations, we usually obtain the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) separately. The derivative of y with respect to x, denoted as \( \frac{dy}{dx} \), then involves a combination of these two derivatives using the chain rule. This is important because it allows us to study how y changes in relation to x, even when both are changing with respect to t.

In applications, parametric differentiation can help in understanding the relationship between different physical quantities that may change with respect to time or other common parameters.

Integration by Substitution

Integration by substitution, also known as u-substitution, is a technique used in integral calculus to find the integral of compound functions. Essentially, it is the reverse process of the chain rule for derivatives.
The idea is to substitute a part of the integral with a new variable (usually u) to simplify the integral. After substituting, we find the integral with respect to u, and then replace u with the original expression. This method is particularly useful when dealing with integrals involving trigonometric functions, logarithms, or other expressions where direct integration is not straightforward.

The second part of the exercise showcases integration by substitution where integration and differentiation involve a more complex expression involving a natural logarithm. Understanding this concept helps with solving more complex integrals that one might encounter in advanced calculus problems.