Question

A particle is moving along the curve \(y=\frac34x^4+3\). The point on the curve at which y-coordinate is changing thrice as fast as the x coordinate, is :

• A. \((1, -\frac{9}{4})\)
• B. \((0, 3)\)
• C. \((1, \frac{15}{4})\)
• D. \((-1, \frac{15}{4})\)

Answer 1

The Correct Option is C

To find the point where the y-coordinate is changing thrice as fast as the x-coordinate, we need to analyze the relationship between the derivatives of y and x with respect to time.
Given the curve:\(y=\frac{3}{4}x^4+3\)

To find the derivative \(\frac{dy}{dx}\), differentiate the equation with respect to \(x\):
\(\frac{dy}{dx}=3x^3\)

We are told that the y-coordinate changes thrice as fast as the x-coordinate, i.e.,\(\frac{dy}{dt}=3\frac{dx}{dt}\)

Using the chain rule, we know:\(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\)

Equating:
\(\frac{dy}{dx}\cdot\frac{dx}{dt}=3\frac{dx}{dt}\)

Since \(\frac{dx}{dt}\neq0\), we can divide both sides by \(\frac{dx}{dt}\):
\(\frac{dy}{dx}=3\)

Substituting the value of \(\frac{dy}{dx}\):
\(3x^3=3\)

Simplifying gives:
\(x^3=1\)
\(x=1\)

Substituting \(x=1\) into the original equation to find \(y\):
\(y=\frac{3}{4}(1)^4+3=\frac{3}{4}+3=\frac{3}{4}+\frac{12}{4}=\frac{15}{4}\)

Thus, the point on the curve where the y-coordinate is changing thrice as fast as the x-coordinate is:
\((1, \frac{15}{4})\)